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All questions of Motion in a Straight Line for NEET Exam
A man starts from his home with a speed of 4 km/h on a straight road up to his office 5 km away and returns to home, then the path length covered is:
- a)-5 km
- b)5 km
- c)0
- d)10 km
Correct answer is option 'D'. Can you explain this answer?
A man starts from his home with a speed of 4 km/h on a straight road up to his office 5 km away and returns to home, then the path length covered is:
a)
-5 km
b)
5 km
c)
0
d)
10 km
|
Preeti Khanna answered |
5 km is the distance covered while going and 5 km while coming back.
⇒ Total path travelled = 10 km
⇒ Total path travelled = 10 km
A particle is moving along a circle of radius R such that it completes one revolution in 40 seconds. What will be the displacement after 2 minutes 20 seconds?
- a)6R
- b)2R
- c)4R
- d)R
Correct answer is option 'B'. Can you explain this answer?
A particle is moving along a circle of radius R such that it completes one revolution in 40 seconds. What will be the displacement after 2 minutes 20 seconds?
a)
6R
b)
2R
c)
4R
d)
R
|
Bhargavi Bajaj answered |
Given:
- Particle is moving along a circle of radius R
- It completes one revolution in 40 seconds
- Time = 2 minutes 20 seconds = 140 seconds
To find: Displacement of the particle after 2 minutes 20 seconds
Solution:
- One revolution is completed in 40 seconds
- Therefore, in 1 second, the particle covers 1/40th of the circumference of the circle
- Circumference of a circle = 2πR
- Therefore, in 1 second, the particle covers 2πR/40 distance
- In 140 seconds, the particle covers (2πR/40) x 140 distance
- Simplifying, we get the displacement as (7πR/2) or 3.5πR
- Option B says the displacement is 2R
- This is not equal to (7πR/2) or 3.5πR
- Therefore, option B is incorrect
- Option A says the displacement is 6R
- This is not equal to (7πR/2) or 3.5πR
- Therefore, option A is incorrect
- Option C says the displacement is 4R
- This is not equal to (7πR/2) or 3.5πR
- Therefore, option C is incorrect
- Option D says the displacement is R
- This is not equal to (7πR/2) or 3.5πR
- Therefore, option D is incorrect
- Hence, the correct answer is option B, which says the displacement is 2R.
- Particle is moving along a circle of radius R
- It completes one revolution in 40 seconds
- Time = 2 minutes 20 seconds = 140 seconds
To find: Displacement of the particle after 2 minutes 20 seconds
Solution:
- One revolution is completed in 40 seconds
- Therefore, in 1 second, the particle covers 1/40th of the circumference of the circle
- Circumference of a circle = 2πR
- Therefore, in 1 second, the particle covers 2πR/40 distance
- In 140 seconds, the particle covers (2πR/40) x 140 distance
- Simplifying, we get the displacement as (7πR/2) or 3.5πR
- Option B says the displacement is 2R
- This is not equal to (7πR/2) or 3.5πR
- Therefore, option B is incorrect
- Option A says the displacement is 6R
- This is not equal to (7πR/2) or 3.5πR
- Therefore, option A is incorrect
- Option C says the displacement is 4R
- This is not equal to (7πR/2) or 3.5πR
- Therefore, option C is incorrect
- Option D says the displacement is R
- This is not equal to (7πR/2) or 3.5πR
- Therefore, option D is incorrect
- Hence, the correct answer is option B, which says the displacement is 2R.
The ratio of magnitude of velocity and speed is always:- a)Equal to or less than one
- b)Less than one
- c)One
- d)Equal to or greater than one
Correct answer is option 'A'. Can you explain this answer?
The ratio of magnitude of velocity and speed is always:
a)
Equal to or less than one
b)
Less than one
c)
One
d)
Equal to or greater than one
|
Anjana Sharma answered |
The displacement of the body in given time is always equal to or less than distance covered, because, velocity is displacement per unit time and speed is distance covered per unit time, therefore, ratio of magnitude of velocity and speed is always equal to or less than one.
Two stones of different masses are thrown vertically upward with same initial velocity. Which one will rise to a greater height ?- a)Stone of lower mass
- b)Stone of higher mass
- c)Both will rise to the same height
- d)Information given is insufficient
Correct answer is option 'C'. Can you explain this answer?
Two stones of different masses are thrown vertically upward with same initial velocity. Which one will rise to a greater height ?
a)
Stone of lower mass
b)
Stone of higher mass
c)
Both will rise to the same height
d)
Information given is insufficient
|
Sushil Kumar answered |
If we write the equation of motion for both the masses and derive a simple equation to find the maximum vertical height let say H, we get for any particle for some mass m,
H = v2 / 2a
Where a is gravitational acceleration and v is vertical speed. Thus as H does not depend upon m, we get both particles would have equal maximum height.
H = v2 / 2a
Where a is gravitational acceleration and v is vertical speed. Thus as H does not depend upon m, we get both particles would have equal maximum height.
Which of the following is not an example of a rectilinear motion?
- a)A car moving on a straight road.
- b)A rain drop falling towards the earth.
- c)A car moving in a circular path.
- d)A wooden block sliding down the inclined plane.
Correct answer is option 'A'. Can you explain this answer?
Which of the following is not an example of a rectilinear motion?
a)
A car moving on a straight road.
b)
A rain drop falling towards the earth.
c)
A car moving in a circular path.
d)
A wooden block sliding down the inclined plane.
|
Neha Joshi answered |
Rectilinear motion is another name for straight-line motion.
A body is said to experience rectilinear motion if any two particles of the body travel the same distance along two parallel straight lines.
A body is said to experience rectilinear motion if any two particles of the body travel the same distance along two parallel straight lines.
A car moving in a circular path changes its direction at every instant. Thus, it is not the rectilinear motion.
36 km/h is equal to:
- a)129. 6 ms-1
- b)600 ms-1
- c)10 ms-1
- d)12.96 ms-1
Correct answer is option 'C'. Can you explain this answer?
36 km/h is equal to:
a)
129. 6 ms-1
b)
600 ms-1
c)
10 ms-1
d)
12.96 ms-1
|
Geetika Shah answered |
1 km = 1000 m and 1 hr = 3600 sec.
Therefore,
If a body does not change its direction during the course of its motion, then ______.
- a)Displacement is greater than path length.
- b)Path length and displacement are equal.
- c)Path length is greater than displacement.
- d)Path length is either greater or equal to displacement.
Correct answer is option 'B'. Can you explain this answer?
If a body does not change its direction during the course of its motion, then ______.
a)
Displacement is greater than path length.
b)
Path length and displacement are equal.
c)
Path length is greater than displacement.
d)
Path length is either greater or equal to displacement.
|
Rajeev Saxena answered |
Since the object does not change its direction it means that the object is traveling in a straight line.
⇒ The path length and displacement will be equal.
⇒ The path length and displacement will be equal.
The numerical ratio of velocity to speed is:- a)Either less than or equal to 1
- b)Equal to 1
- c)Greater than 1
- d)Less than 1
Correct answer is option 'A'. Can you explain this answer?
The numerical ratio of velocity to speed is:
a)
Either less than or equal to 1
b)
Equal to 1
c)
Greater than 1
d)
Less than 1
|
|
Charvi Mehta answered |
When an object is moving along a straight path, magnitude of average velocity is equal to the average speed. Therefore, numerical ratio of average velocity to average speed is one in a straight line motion.
But, during curved motion, the displacement<distance covered, so the velocity<speed.
A vehicle travels half the distance L with speed V1and the other half with speed V2, then its average speed is
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A vehicle travels half the distance L with speed V1and the other half with speed V2, then its average speed is
a)
b)
c)
d)
|
|
Gaurav Kumar answered |
Let the vehicle travels from A to B. Distances, velocities, and time taken are shown. To calculate average speed we will calculate the total distance covered and will divide it by the time interval in which it covers that total distance.
Time taken to travel first half distance t1
Time taken to travel first half distance t1
Time taken to travel first half distance t1
Total time taken = t1 + t2
We know that vav = Average speed = Total Distance / Total Time
If the signs of the velocity and acceleration of a particle are the same, the speed of the particle ______________.- a)Remains same
- b)Increases
- c)Decreases
- d)Cannot conclude anything
Correct answer is option 'B'. Can you explain this answer?
If the signs of the velocity and acceleration of a particle are the same, the speed of the particle ______________.
a)
Remains same
b)
Increases
c)
Decreases
d)
Cannot conclude anything
|
|
Pooja Shah answered |
In this question, it has been asked about the signs of velocity and acceleration not about the values of velocity and acceleration.
So, if both the entities have same the signs, then the speed of the object will increase.
If it is accelerating in the positive direction, the negative velocity is decreasing, and the object is slowing down.
A truck has a velocity of 2 m /s at time t=0. It accelerates at 2 m / s2 on seeing police .What is its velocity in m/s at a time of 2 sec- a)6
- b)7
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
A truck has a velocity of 2 m /s at time t=0. It accelerates at 2 m / s2 on seeing police .What is its velocity in m/s at a time of 2 sec
a)
6
b)
7
c)
3
d)
4
|
|
Naina Sharma answered |
Explanation:
Initial velocity u = 2 m/s
final velocity = v m/s
Time duration = final time - initial time = 2-0 = 2 s
acceleration a = 2 m/s2
We know,
v = u + at
=> v = 2+2x2
=> v = 6 m/s
A car covers a distance of 5 km in 5 mins, its average speed is equal to- a)1 km ⁄ h
- b)25 km ⁄ h
- c)60 km ⁄ h
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
A car covers a distance of 5 km in 5 mins, its average speed is equal to
a)
1 km ⁄ h
b)
25 km ⁄ h
c)
60 km ⁄ h
d)
None of the above
|
|
Raghav Bansal answered |
5 km in 5 mins, which means 1 km in 1 min
Hence, it will travel 60 km in 60 mins, i.e. speed = 60km/hr
Hence, it will travel 60 km in 60 mins, i.e. speed = 60km/hr
The ratio of displacement to distance is- a)Greater than 1
- b)Either less than or equal to 1
- c)Less than 1
- d)Equal to 1
Correct answer is option 'B'. Can you explain this answer?
The ratio of displacement to distance is
a)
Greater than 1
b)
Either less than or equal to 1
c)
Less than 1
d)
Equal to 1
|
|
Saumya Dey answered |
Ratio of Displacement to Distance
Displacement and distance are two terms used in physics to describe the motion of an object. Displacement refers to the change in position of an object from its initial to final position, whereas distance refers to the total path covered by the object during its motion.
Ratio of Displacement to Distance Formula
The ratio of displacement to distance is given by:
Displacement / Distance
This ratio is used to determine the efficiency of an object's motion. If the ratio is equal to 1, it means that the object has moved in a straight line from its initial to final position. If the ratio is less than 1, it means that the object has taken a longer path to reach its final position, and if the ratio is greater than 1, it means that the object has moved back and forth before reaching its final position.
Answer Explanation
The correct answer to the question is option B, which states that the ratio of displacement to distance is either less than or equal to 1. This is because an object can never cover a greater distance than its displacement. The displacement represents the shortest distance between the initial and final positions of an object, whereas the distance represents the total path covered by the object.
Therefore, the displacement can be equal to or less than the distance covered by the object, but it can never be greater than the distance. This is the reason why the ratio of displacement to distance is either less than or equal to 1.
Displacement and distance are two terms used in physics to describe the motion of an object. Displacement refers to the change in position of an object from its initial to final position, whereas distance refers to the total path covered by the object during its motion.
Ratio of Displacement to Distance Formula
The ratio of displacement to distance is given by:
Displacement / Distance
This ratio is used to determine the efficiency of an object's motion. If the ratio is equal to 1, it means that the object has moved in a straight line from its initial to final position. If the ratio is less than 1, it means that the object has taken a longer path to reach its final position, and if the ratio is greater than 1, it means that the object has moved back and forth before reaching its final position.
Answer Explanation
The correct answer to the question is option B, which states that the ratio of displacement to distance is either less than or equal to 1. This is because an object can never cover a greater distance than its displacement. The displacement represents the shortest distance between the initial and final positions of an object, whereas the distance represents the total path covered by the object.
Therefore, the displacement can be equal to or less than the distance covered by the object, but it can never be greater than the distance. This is the reason why the ratio of displacement to distance is either less than or equal to 1.
A ball is thrown vertically upward. At the highest point in its path, which of the following statements is correct?- a)The velocity is zero and the acceleration is pointing down
- b)The velocity and acceleration are both zero
- c)The acceleration is zero and the velocity is pointing up
- d)The acceleration is zero and the velocity is pointing down
Correct answer is option 'A'. Can you explain this answer?
A ball is thrown vertically upward. At the highest point in its path, which of the following statements is correct?
a)
The velocity is zero and the acceleration is pointing down
b)
The velocity and acceleration are both zero
c)
The acceleration is zero and the velocity is pointing up
d)
The acceleration is zero and the velocity is pointing down
|
|
Geetika Shah answered |
Until and unless the velocity of any object comes to zero, any peek in its displacement-time graph can’t be achieved. Which means for the height to be maximised the velocity must become zero. And any time of this motion the acceleration is always a constant downward due to gravity.
A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed v. The average speed is 48 km/h. Find the value of v- a)56 km/h
- b)60 km/h [1991]
- c)50 km/h
- d)48 km/h
Correct answer is option 'B'. Can you explain this answer?
A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed v. The average speed is 48 km/h. Find the value of v
a)
56 km/h
b)
60 km/h [1991]
c)
50 km/h
d)
48 km/h
|
Naveen Menon answered |
or v = 60 km/h
⇒ v = 60 km/h]
On velocity-time graph when two curves coincide, the objects will have- a)Same acceleration
- b)Same velocity
- c)Different velocity
- d)Different acceleration
Correct answer is option 'B'. Can you explain this answer?
On velocity-time graph when two curves coincide, the objects will have
a)
Same acceleration
b)
Same velocity
c)
Different velocity
d)
Different acceleration
|
|
Anjana Sharma answered |
On velocity-time graph, the point at which two curves coincide will indicate that both the objects have same velocity at that instant.
On acceleration-time graph, a horizontal line indicates:- a)zero displacement
- b)zero acceleration
- c)constant velocity
- d)constant acceleration
Correct answer is option 'D'. Can you explain this answer?
On acceleration-time graph, a horizontal line indicates:
a)
zero displacement
b)
zero acceleration
c)
constant velocity
d)
constant acceleration
|
|
Anjana Sharma answered |
A horizontal line on a acceleration- time graph indicates that object has constant acceleration.
The dimensions of (change in velocity / time) is- a)[LT-2]
- b)[LT-1]
- c)[L/T-1]
- d)[L/T-2]
Correct answer is option 'A'. Can you explain this answer?
The dimensions of (change in velocity / time) is
a)
[LT-2]
b)
[LT-1]
c)
[L/T-1]
d)
[L/T-2]
|
|
Naina Sharma answered |
We know that change in velocity / time is acceleration.
And dimension of acceleration is LT-2
A boy throws up a ball in a stationary lift and the ball returns to his hands in 10 s. Now if the lift starts moving up at a speed of 5 m/s. The time taken for a ball thrown straight up to return to his hands is:- a)more than 10 s
- b)Less than 10 s
- c)insufficient information given
- d)equal to 10 s
Correct answer is option 'D'. Can you explain this answer?
A boy throws up a ball in a stationary lift and the ball returns to his hands in 10 s. Now if the lift starts moving up at a speed of 5 m/s. The time taken for a ball thrown straight up to return to his hands is:
a)
more than 10 s
b)
Less than 10 s
c)
insufficient information given
d)
equal to 10 s
|
|
Suresh Iyer answered |
Lift moving with Uniform speed = 5m / s
Ball is thrown inside the lift
The uniform velocity of the lift does not affect the relative velocity of the ball with respect to the boy.
Relative velocity of ball = 49m/s
⇒ Relative acceleration = 9.8
Thus, u(relative) = velocity of bell
t = 2ur / gr
= (2 × 49) / 9.8
⇒ t = 10s
so, the ball will still return in 10 secs
The dimensions of instantaneous acceleration is:- a)[ LT-2]
- b)[L/T2]
- c)[L/T-1]
- d)[LT-1]
Correct answer is option 'A'. Can you explain this answer?
The dimensions of instantaneous acceleration is:
a)
[ LT-2]
b)
[L/T2]
c)
[L/T-1]
d)
[LT-1]
|
|
Neha Joshi answered |
Instantaneous acceleration is rate of change of instantaneous velocity with time or velocity upon total time, hence we get its dimension is same as of acceleration i.e. LT-2
What does this graph indicate about the motion of an object?
- a)Object is moving in positive direction with positive acceleration
- b)Object is moving in negative direction with negative acceleration
- c)An object is moving in positive direction till time t1, and then turns back with the same Velocity
- d)Object is moving in positive direction with negative acceleration
Correct answer is option 'D'. Can you explain this answer?
What does this graph indicate about the motion of an object?
a)
Object is moving in positive direction with positive acceleration
b)
Object is moving in negative direction with negative acceleration
c)
An object is moving in positive direction till time t1, and then turns back with the same Velocity
d)
Object is moving in positive direction with negative acceleration
|
Top Rankers answered |
Through the given graph we get that:
Velocity is decreasing from a positive constant to some negative number with zero at t1
Velocity is decreasing from a positive constant to some negative number with zero at t1
Distance from the original point increases from zero to t1 and then it decreases afterwards.
The object is moving in a positive direction at first and after time t1 it moves in a negative direction.
What does this graph indicate about the motion of an object ?
- a)object is moving in negative direction with positive acceleration
- b)object is moving in negative direction with negative acceleration
- c)object is moving in positive direction with negative acceleration
- d)object is moving in positive direction with positive acceleration
Correct answer is option 'C'. Can you explain this answer?
What does this graph indicate about the motion of an object ?
a)
object is moving in negative direction with positive acceleration
b)
object is moving in negative direction with negative acceleration
c)
object is moving in positive direction with negative acceleration
d)
object is moving in positive direction with positive acceleration
|
|
Raghav Bansal answered |
Velocity is decreasing so, acceleration is negative. Velocity is positive so direction is positive.
Hence option (C) is correct.
If the position- time graph is a straight line parallel to the time axis- a)The velocity is zero
- b)The velocity is decreasing
- c)The velocity is increasing
- d)The velocity is constant but non zero
Correct answer is option 'A'. Can you explain this answer?
If the position- time graph is a straight line parallel to the time axis
a)
The velocity is zero
b)
The velocity is decreasing
c)
The velocity is increasing
d)
The velocity is constant but non zero
|
|
Neha Sharma answered |
Explanation:Position-time graph of horizontal straight line parallel to time axis represents that the position of the body does not changes with the passage of time. So it represents the rest state of motion.It means velocity of object is zero.
The position x of a particle varies with time (t) as x = 3t2 – 2t3 .The acceleration of the particle will be zero at time- a)1
- b)3/2
- c)0
- d)1/2
Correct answer is option 'D'. Can you explain this answer?
The position x of a particle varies with time (t) as x = 3t2 – 2t3 .The acceleration of the particle will be zero at time
a)
1
b)
3/2
c)
0
d)
1/2
|
Lavanya Menon answered |
We know that acceleration is the second derivative of change in displacement wrt time, hence we get a = 6 - 12t
When a = 0 = 6 -12t ;
We get t = ½
When a = 0 = 6 -12t ;
We get t = ½
Suppose our school is 1 km away from our house, and we go to school in the morning, and in the afternoon we come back. Then, the total displacement for the entire trip is- a)1 km
- b)2 km
- c)-1 km
- d)0 km
Correct answer is option 'D'. Can you explain this answer?
Suppose our school is 1 km away from our house, and we go to school in the morning, and in the afternoon we come back. Then, the total displacement for the entire trip is
a)
1 km
b)
2 km
c)
-1 km
d)
0 km
|
|
Lavanya Menon answered |
The displacement is 0 because the initial and final position are the same, i.e. the home, and hence, displacement is 0.
Two balls of equal mass and of Perfectly inelastic material are lying on the floor. One of the balls with velocity V is made to strike the second ball. Both the balls after impact will move with a velocity
- a)v
- b)v/8
- c)v/4
- d)v/2
Correct answer is option 'D'. Can you explain this answer?
Two balls of equal mass and of Perfectly inelastic material are lying on the floor. One of the balls with velocity V is made to strike the second ball. Both the balls after impact will move with a velocity
a)
v
b)
v/8
c)
v/4
d)
v/2
|
|
Neha Joshi answered |
Given:
Let two balls A and B have mass mA, mB respectively, and their initial velocities are uA and uB. After the collision, they will move with the same velocity, vo.
Given that mass of both balls are same.
So mA = mB = m
uA = V, uB = 0
From the Concept of Momentum Conservation:
mAuA + mBuB = (m+m)vo
mV = 2mvo
vo = V/2
Both the balls after impact will move with velocity v/2.
The velocity of the particle at any time t is given by v = 2t(3 − t)m s−1. At what time is its velocity maximum?- a)2 s
- b)3 s
- c)2/3s
- d)3/2 s
Correct answer is option 'D'. Can you explain this answer?
The velocity of the particle at any time t is given by v = 2t(3 − t)m s−1. At what time is its velocity maximum?
a)
2 s
b)
3 s
c)
2/3s
d)
3/2 s
|
Hansa Sharma answered |
given v = 2t(3 − t) or v = 6t - 2t2
For V maximum or minimum
The acceleration at any instant is the slope of the tangent of the ________ curve at that instant:- a)a-t
- b)v-t
- c)x-v
- d)x-t
Correct answer is option 'B'. Can you explain this answer?
The acceleration at any instant is the slope of the tangent of the ________ curve at that instant:
a)
a-t
b)
v-t
c)
x-v
d)
x-t
|
Om Desai answered |
This can be verified graphically
ObtaIning instantaneous acceleration from graph
Figure shows the displacement – time graph of a particle moving along X – axis.
- a)particle moves at a variable velocity
- b)particle moves at a constant velocity upto a time t0 and then stops
- c)particle moves at a constant velocity upto a time t0 and then velocity increases
- d)particle moves at increasing velocity upto t0 and then velocity becomes constant
Correct answer is option 'B'. Can you explain this answer?
Figure shows the displacement – time graph of a particle moving along X – axis.
a)
particle moves at a variable velocity
b)
particle moves at a constant velocity upto a time t0 and then stops
c)
particle moves at a constant velocity upto a time t0 and then velocity increases
d)
particle moves at increasing velocity upto t0 and then velocity becomes constant
|
|
Preeti Iyer answered |
Because velocity is the slope of the x-t graph.(dx/dt=V) and by looking at the graph it can be seen that upto time T0 slope is constant so velocity will be constant and after time T0 slope is zero ( line is parallel to X axis) so the velocity will be zero.
A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h1, h2 and h3 is [NEET 2013]- a)
- b)h2 = 3h1 and h3 = 3h2
- c)h1 = h2 = h3
- d)h1 = 2h2 = 3h3
Correct answer is option 'A'. Can you explain this answer?
A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h1, h2 and h3 is [NEET 2013]
a)
b)
h2 = 3h1 and h3 = 3h2
c)
h1 = h2 = h3
d)
h1 = 2h2 = 3h3
|
|
Geetika Shah answered |
⇒ h2 = 375
An object is said to be in uniform motion in a straight line if its displacement- a)is decreasing in equal intervals of time
- b)is increasing in equal intervals of time
- c)is equal in equal intervals of time.
- d)is equal in not equal intervals of time.
Correct answer is option 'C'. Can you explain this answer?
An object is said to be in uniform motion in a straight line if its displacement
a)
is decreasing in equal intervals of time
b)
is increasing in equal intervals of time
c)
is equal in equal intervals of time.
d)
is equal in not equal intervals of time.
|
Mira Joshi answered |
Explanation:Uniform motion is the kind of motion in which a body covers equal displacement in equal intervals of time. It does not matter how small the time intervals are, as long as the displacements covered are equal.
If a body is involved in rectilinear motion and the motion is uniform, then the acceleration of the body must be zero.
Rain falls at a speed of 50m/s and a child walks on a straight road from east to west at a speed of 100m/s. To find the direction in which the child should hold the umbrella use- a)resultant speed of child and rain
- b)resultant speed of rain and child
- c)relative speed of rain with respect to child
- d)relative speed of child with respect to rain
Correct answer is option 'C'. Can you explain this answer?
Rain falls at a speed of 50m/s and a child walks on a straight road from east to west at a speed of 100m/s. To find the direction in which the child should hold the umbrella use
a)
resultant speed of child and rain
b)
resultant speed of rain and child
c)
relative speed of rain with respect to child
d)
relative speed of child with respect to rain
|
|
Rajesh Gupta answered |
The child should hold the umbrella in the direction at which the rain appears to fall on him, which is the direction of relative speed of rain with respect to the child.
Speed time graph of a particle moving along a fixed direction is shown in the figure below. The average speed of the particle over the interval: t = 0 s to 10 s.
- a)16 m/s
- b)6 m/s
- c)10 m/s
- d)0.6 m/s
Correct answer is option 'B'. Can you explain this answer?
Speed time graph of a particle moving along a fixed direction is shown in the figure below. The average speed of the particle over the interval: t = 0 s to 10 s.
a)
16 m/s
b)
6 m/s
c)
10 m/s
d)
0.6 m/s
|
|
Om Desai answered |
From the given graph we get that the area of the given graph is total distance covered that is
= ½ x 10 x 12
= 60m
And total time taken is 10 sec
Thus average speed is 60m / 10sec
= 6 m/s
= ½ x 10 x 12
= 60m
And total time taken is 10 sec
Thus average speed is 60m / 10sec
= 6 m/s
Jagadeesh, on driving his way to school, calculates the speed for the trip to be 20 km/hr. After reaching the school he found the school was closed. So he immediately started returning home. While on his return trip, due to less traffic, he calculates the speed to be 40 km/hr. Calculate Jagadeesh's average speed for the entire journey.
- a)26.67 kmhr-1
- b)33.34 kmhr-1
- c)50 kmhr-1
- d)30 kmhr-1
Correct answer is option 'A'. Can you explain this answer?
Jagadeesh, on driving his way to school, calculates the speed for the trip to be 20 km/hr. After reaching the school he found the school was closed. So he immediately started returning home. While on his return trip, due to less traffic, he calculates the speed to be 40 km/hr. Calculate Jagadeesh's average speed for the entire journey.
a)
26.67 kmhr-1
b)
33.34 kmhr-1
c)
50 kmhr-1
d)
30 kmhr-1
|
Ciel Knowledge answered |
Let t1 and t2 be the time taken for Jagadeesh to go to school and then back home. If s is the displacement, then
Total time taken
Total displacement=2 s
Total time taken
Total displacement=2 s
A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is [2011]- a)1 m/s2
- b)6 m/s2
- c)7 m/s2
- d)5 m/s2
Correct answer is option 'D'. Can you explain this answer?
A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is [2011]
a)
1 m/s2
b)
6 m/s2
c)
7 m/s2
d)
5 m/s2
|
|
Rohit Jain answered |
Average acceleration
< a > = 5 m/sec2
Two escalators move people up and down between floors of a shopping mall at the same rate, either up or down. What of the following statements regarding the motion of the escalators is true?- a)The escalators have different speeds and velocities.
- b)The escalators have the same velocity, but different speeds.
- c)The escalators have the same speed and velocity.
- d)The escalators have the same speed, but different velocities.
Correct answer is option 'D'. Can you explain this answer?
Two escalators move people up and down between floors of a shopping mall at the same rate, either up or down. What of the following statements regarding the motion of the escalators is true?
a)
The escalators have different speeds and velocities.
b)
The escalators have the same velocity, but different speeds.
c)
The escalators have the same speed and velocity.
d)
The escalators have the same speed, but different velocities.
|
|
Riya Banerjee answered |
As the particles are moving at the same rate, their speed is the same but velocities will be different beacuse their directions will be different.
The motion of a freely falling body is an example of:- a)uniformly accelerated motion
- b)non uniformly accelerated motion
- c)uniform motion
- d)none of the above
Correct answer is option 'A'. Can you explain this answer?
The motion of a freely falling body is an example of:
a)
uniformly accelerated motion
b)
non uniformly accelerated motion
c)
uniform motion
d)
none of the above
|
|
Om Desai answered |
A freely falling body observes a constant downwards force of gravity which hence applies a constant downward acceleration upon the body.
The distance travelled by a particle starting from rest and moving with an acceleration 
, in the third second is: [2008]- a)6 m
- b)4 m
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The distance travelled by a particle starting from rest and moving with an acceleration , in the third second is: [2008]
, in the third second is: [2008]a)
6 m
b)
4 m
c)
d)
|
Dipanjan Mehta answered |
Distance travelled in the nth second is given by
put u = 0, 
A stone released with zero velocity from the top of a tower, reaches the ground in 4 sec. The height of the tower is (g = 10m / s2 ) [1995]- a)20 m
- b)40 m
- c)80 m
- d)160 m
Correct answer is option 'C'. Can you explain this answer?
A stone released with zero velocity from the top of a tower, reaches the ground in 4 sec. The height of the tower is (g = 10m / s2 ) [1995]
a)
20 m
b)
40 m
c)
80 m
d)
160 m
|
Ashwini Khanna answered |
Initial velocity (u) = 0; Time (t) = 4 sec and gravitational acceleration (g) = 10 m/s2.
Height of tower
Height of tower
= 80 m.
The position x of a particle with respect to time t along x-axis is given by x = 9t2 −t3. where x is in metres and t is in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction?- a)54 m
- b)81 m
- c)24 m
- d)32 m
Correct answer is option 'A'. Can you explain this answer?
The position x of a particle with respect to time t along x-axis is given by x = 9t2 −t3. where x is in metres and t is in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction?
a)
54 m
b)
81 m
c)
24 m
d)
32 m
|
|
Anjali Sharma answered |
Given: x = 9t2 − t3
for Maximum speed, dv/dt
= 0 ⇒ 18 - 6t = 0
∴ t = 3s.
∴ xmax = 81m - 27m
= 54 m. (From x = 9t2 - t3)
= 0 ⇒ 18 - 6t = 0
∴ t = 3s.
∴ xmax = 81m - 27m
= 54 m. (From x = 9t2 - t3)
The displacement of a particle along x-axis is given by x = 4 + 6t + 5t2. Its acceleration at t = 2s- a)2 m/s2
- b)10 m/s2
- c)1000 m/s2
- d)100 m/s2
Correct answer is option 'B'. Can you explain this answer?
The displacement of a particle along x-axis is given by x = 4 + 6t + 5t2. Its acceleration at t = 2s
a)
2 m/s2
b)
10 m/s2
c)
1000 m/s2
d)
100 m/s2
|
|
Mira Sharma answered |
usingdifferentiation
x = 18t + 5t^2
dx/dt = 18 + 10t
v = 18 + 10t
at t = 2s
v = 18 + 10x2
= 18 + 20 =38m/s
at t = 3s
v = 18 + 10x3
= 18 + 20 = 48m/s
vav= 38+48/2= 43m/s
dv/dt = 10
a = 10
a =10m/s^2
On acceleration-time graph the area under the curve equals the:- a)Meaningless
- b)Change in velocity
- c)Displacement
- d)Change in acceleration
Correct answer is option 'B'. Can you explain this answer?
On acceleration-time graph the area under the curve equals the:
a)
Meaningless
b)
Change in velocity
c)
Displacement
d)
Change in acceleration
|
|
Nandini Patel answered |
The area under an acceleration graph represents the change in velocity. In other words, the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval.
Slope of displacement time graph or x-t graph gives us the particles’ ____________.- a)displacement
- b)deceleration
- c)velocity
- d)acceleration
Correct answer is option 'C'. Can you explain this answer?
Slope of displacement time graph or x-t graph gives us the particles’ ____________.
a)
displacement
b)
deceleration
c)
velocity
d)
acceleration
|
|
Rahul Bansal answered |
Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value (magnitude) of velocity is called "speed", being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s) or as the SI base unit of (m⋅s−1).
A body of mass 1kg and 10kg are dropped simultaneously from the top of a tower. The ratio of the time taken by them to reach the ground is- a)9.8:98
- b)1:1
- c)10:1
- d)1:10
Correct answer is option 'B'. Can you explain this answer?
A body of mass 1kg and 10kg are dropped simultaneously from the top of a tower. The ratio of the time taken by them to reach the ground is
a)
9.8:98
b)
1:1
c)
10:1
d)
1:10
|
|
Anjana Sharma answered |
1:1, the time taken does not depend on the mass of the falling object.
A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is (g = 10 m/s2) [1991]- a)60 m
- b)45 m
- c)80 m
- d)50 m
Correct answer is option 'B'. Can you explain this answer?
A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is (g = 10 m/s2) [1991]
a)
60 m
b)
45 m
c)
80 m
d)
50 m
|
Shounak Nair answered |
Let h = the height of the tower that needs to be determined.
Let t be the time of fall.
Then (t - 2) would be the time to reach the top of the 40 meter mark, and let d be the distance fallen till it reaches the 40 m mark.
Using kinematics, we can write: d = 1/2 g (t-2)^2
And H = 1/2 g t^2
H also = 40 + d
Then: H = 40 + 1/2 g (t-2)^2 = 1/2 g t^2
Expand: 40 +5 (t^2 - 4 t + 4) = 5 t^2
40 + 5 t^2 - 20 t + 20 = 5 t^2
Add like terms: - 20 t = - 60
t = 3 s
(t-2) = 1 second
In one second an object falls 5 m
Then H = 45 m
Let t be the time of fall.
Then (t - 2) would be the time to reach the top of the 40 meter mark, and let d be the distance fallen till it reaches the 40 m mark.
Using kinematics, we can write: d = 1/2 g (t-2)^2
And H = 1/2 g t^2
H also = 40 + d
Then: H = 40 + 1/2 g (t-2)^2 = 1/2 g t^2
Expand: 40 +5 (t^2 - 4 t + 4) = 5 t^2
40 + 5 t^2 - 20 t + 20 = 5 t^2
Add like terms: - 20 t = - 60
t = 3 s
(t-2) = 1 second
In one second an object falls 5 m
Then H = 45 m
Linear inequalities are graphically represented on Cartesian plane by a- a)negative full space
- b)closed half space
- c)open half space
- d)positive full space
- e)
Correct answer is option 'B'. Can you explain this answer?
Linear inequalities are graphically represented on Cartesian plane by a
a)
negative full space
b)
closed half space
c)
open half space
d)
positive full space
e)
|
|
Muskaan Kumar answered |
The graph of an inequality in two variables is the set of points that represents all solutions to the inequality. A linear inequality divides the coordinate plane into two halves by a boundary line where one half represents the solutions of the inequality. The boundary line is dashed for > and < and solid for ≤ and ≥.
The kinematic equation of motion v = u+at is not applicable if :- a)Displacement is not constant
- b)motion is non-uniform.
- c)acceleration is not constant
- d)velocity is not constant
Correct answer is option 'C'. Can you explain this answer?
The kinematic equation of motion v = u+at is not applicable if :
a)
Displacement is not constant
b)
motion is non-uniform.
c)
acceleration is not constant
d)
velocity is not constant
|
Yash Ghoshal answered |
Explanation:
The kinematic equation of motion v = u + at is one of the four equations of motion which are used to describe the motion of an object under constant acceleration. The equation shows the relationship between the initial velocity (u), acceleration (a), time (t), and final velocity (v) of an object. However, this equation is not applicable if the acceleration is not constant.
Constant acceleration means that the rate of change of velocity is constant. If the acceleration is not constant, then the rate of change of velocity is changing, and the equation v = u + at does not hold true for the entire motion.
For example, if an object is moving in a circular path, its acceleration changes direction at every point, and hence, the acceleration is not constant. In this case, the kinematic equation cannot be used to describe the motion of the object.
Similarly, if an object is thrown vertically upwards, the acceleration of the object changes as it moves upwards and then downwards due to the force of gravity. Hence, the kinematic equation v = u + at cannot be used to describe the entire motion of the object.
Conclusion:
In conclusion, the kinematic equation of motion v = u + at is not applicable if the acceleration is not constant, and the rate of change of velocity is changing. The equation is valid only for objects that are moving under constant acceleration. Hence, before using this equation to solve any problem, it is important to check whether the acceleration is constant or not.
The kinematic equation of motion v = u + at is one of the four equations of motion which are used to describe the motion of an object under constant acceleration. The equation shows the relationship between the initial velocity (u), acceleration (a), time (t), and final velocity (v) of an object. However, this equation is not applicable if the acceleration is not constant.
Constant acceleration means that the rate of change of velocity is constant. If the acceleration is not constant, then the rate of change of velocity is changing, and the equation v = u + at does not hold true for the entire motion.
For example, if an object is moving in a circular path, its acceleration changes direction at every point, and hence, the acceleration is not constant. In this case, the kinematic equation cannot be used to describe the motion of the object.
Similarly, if an object is thrown vertically upwards, the acceleration of the object changes as it moves upwards and then downwards due to the force of gravity. Hence, the kinematic equation v = u + at cannot be used to describe the entire motion of the object.
Conclusion:
In conclusion, the kinematic equation of motion v = u + at is not applicable if the acceleration is not constant, and the rate of change of velocity is changing. The equation is valid only for objects that are moving under constant acceleration. Hence, before using this equation to solve any problem, it is important to check whether the acceleration is constant or not.
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