All Exams >
A Level >
Mathematics for A Level >
All Questions
All questions of Kinematics of Motion in a Straight Line for A Level 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
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.
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
|
Nandini Patel answered |
1 revolution in 40 seconds
In 1 second it covers = 1/40 revolution
In 140 sec = 1/40 * 140 =7/2 rotation = 3 and half rotation
In 1 second it covers = 1/40 revolution
In 140 sec = 1/40 * 140 =7/2 rotation = 3 and half rotation
Then the particle will be on the diametrically opposite end.
Therefore, Displacement = R + R = 2R
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.
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.
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
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 displacement of a particle starting from rest (at t = 0) is given by s = 6t2-t3 . The time in seconds at which the particle will attain zero velocity again, is
- a)6
- b)4
- c)2
- d)8
Correct answer is option 'B'. Can you explain this answer?
The displacement of a particle starting from rest (at t = 0) is given by s = 6t2-t3 . The time in seconds at which the particle will attain zero velocity again, is
a)
6
b)
4
c)
2
d)
8
|
|
Preeti Iyer answered |
v = ds/dt
v = 12t - 3t2
0 = 3t (4 - t)
⇒ t = 0 or 4
Here t = 0 is not realistic so at t = 4 the velocity will be zero.
Thus, the particle will attain zero velocity at the time of 4 seconds.
⇒ t = 0 or 4
Here t = 0 is not realistic so at t = 4 the velocity will be zero.
Thus, the particle will attain zero velocity at the time of 4 seconds.
A tennis ball is thrown straight up and caught at the same height. Which of the following can describe the motion of the ball when it reaches the apex?- a)The velocity of the ball is zero.
- b)The acceleration of the ball is zero.
- c)The acceleration of the ball is 9.8 m/s2 down
- d)The acceleration of the ball is 9.8 m/s2 up.
Correct answer is option 'A,C'. Can you explain this answer?
A tennis ball is thrown straight up and caught at the same height. Which of the following can describe the motion of the ball when it reaches the apex?
a)
The velocity of the ball is zero.
b)
The acceleration of the ball is zero.
c)
The acceleration of the ball is 9.8 m/s2 down
d)
The acceleration of the ball is 9.8 m/s2 up.
|
Ciel Knowledge answered |
When the ball is at its peak possible position, it's velocity remains Zero. Unfortunately, this moment of zero velocity is hardly visible, since it occurs in a very very short amount of time.
Now if you ask about the acceleration at the peak possible position, it turns out to be very interesting. The acceleration remains 9.81 ms^-2. In fact it is one of the very rare situations where the body is accelerated, still there is an absolute zero velocity.
Now if you ask about the acceleration at the peak possible position, it turns out to be very interesting. The acceleration remains 9.81 ms^-2. In fact it is one of the very rare situations where the body is accelerated, still there is an absolute zero velocity.
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.
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.
The variation of velocity of a particle moving along straight line is shown in the figure. The distance travelled by the particle in 4 s is 
- a)25m
- b)30m
- c)55m
- d)60m
Correct answer is option 'C'. Can you explain this answer?
The variation of velocity of a particle moving along straight line is shown in the figure. The distance travelled by the particle in 4 s is
a)
25m
b)
30m
c)
55m
d)
60m
|
Hansa Sharma answered |
The distance traveled is equal to the area under the curve which is equal to
S = ½ x 20 x1 + 20 x 1 +½ x 10 x1 + 10 x 2
= 10 + 20 + 5 + 20
= 55m
S = ½ x 20 x1 + 20 x 1 +½ x 10 x1 + 10 x 2
= 10 + 20 + 5 + 20
= 55m
A particle is moving so that its displacement s is given as s = t3- 6t2 + 3t + 4 meter. Its velocity at the instant when its acceleration is zero will be-- a)3 m/s
- b)-12 m/s
- c)42 m/s
- d)-9 m/s
Correct answer is option 'D'. Can you explain this answer?
A particle is moving so that its displacement s is given as s = t3- 6t2 + 3t + 4 meter. Its velocity at the instant when its acceleration is zero will be-
a)
3 m/s
b)
-12 m/s
c)
42 m/s
d)
-9 m/s
|
Mira Joshi answered |
For the given particle, v = 3t2 - 12t + 3
And its acceleration, a = 6t - 12
Thus we get a = 0
At t = 2 sec,
Hence v at t = 2 sec
V = 3 x 4 - 12 x 2 + 3
= -9m/s
And its acceleration, a = 6t - 12
Thus we get a = 0
At t = 2 sec,
Hence v at t = 2 sec
V = 3 x 4 - 12 x 2 + 3
= -9m/s
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
A car A is going north-east at 80 km/hr and another car B is going south-east at 60 km/hr. Then the direction of the velocity of A relative to B makes with the north an angle a such that tan a is -- a)1/7
- b)3/4
- c)4/3
- d)3/5
Correct answer is option 'A'. Can you explain this answer?
A car A is going north-east at 80 km/hr and another car B is going south-east at 60 km/hr. Then the direction of the velocity of A relative to B makes with the north an angle a such that tan a is -
a)
1/7
b)
3/4
c)
4/3
d)
3/5
|
|
Preeti Khanna answered |
Both the cars are moving in perpendicular directions and thus we stop b and see speed of a wrt b, we get the direction of a is 80.cos 45 + 60.cos 45 east and 80.sin 45 - 60.sin 45.
That is the directions are 140.cos 45 east and 20.sin 45 north.
Thus the angle with north is tan-1 20/140
= tan-1 1/7
That is the directions are 140.cos 45 east and 20.sin 45 north.
Thus the angle with north is tan-1 20/140
= tan-1 1/7
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.
A car travels from A to B at a speed of 20 km h_1 and returns at a speed of 30 km h_1. The average speed of the car for the whole journey is- a)5 km h_1
- b)24 km h_1
- c) 25 km h_1
- d)50 km h_1
Correct answer is option 'B'. Can you explain this answer?
A car travels from A to B at a speed of 20 km h_1 and returns at a speed of 30 km h_1. The average speed of the car for the whole journey is
a)
5 km h_1
b)
24 km h_1
c)
25 km h_1
d)
50 km h_1
|
|
Naina Sharma answered |
Let distance from A to B be x
Time for travel from A to B= x/20
Time for travel from B to A= x/30
Hence,speed for total distance 2x is:
A particle is fired with velocity v making an angle q with the horizontal. What is the magnitude of change in velocity when it is at the highest point? - a)u cosθ
- b)u
- c)u sinθ
- d)(u cosθ -u)
Correct answer is option 'C'. Can you explain this answer?
A particle is fired with velocity v making an angle q with the horizontal. What is the magnitude of change in velocity when it is at the highest point?
a)
u cosθ
b)
u
c)
u sinθ
d)
(u cosθ -u)
|
Krishna Iyer answered |
velocity at the highest point 
velocity at the starting point 
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.
A projectile of mass 1 kg is projected with a velocity of 
m/s such that it strikes on the same level as the point of projection at a distance of 
m. Which of the following options are incorrect.- a)The maximum height reached by the projectile can be 0.25 m
- b)The minimum velocity during its motion can be
m/s - c)The time taken for the flight can be
sec. - d)Maximum potential energy during its motion can be 6J.
Correct answer is option 'C,D'. Can you explain this answer?
A projectile of mass 1 kg is projected with a velocity of m/s such that it strikes on the same level as the point of projection at a distance of m. Which of the following options are incorrect.
m/s such that it strikes on the same level as the point of projection at a distance of m. Which of the following options are incorrect.a)
The maximum height reached by the projectile can be 0.25 m
b)
The minimum velocity during its motion can be m/s
m/sc)
The time taken for the flight can be sec.
sec.d)
Maximum potential energy during its motion can be 6J.
|
Kamna Science Academy answered |
R=u2sin2θ=√3
(√20)2sin2θ/10=√3
sin2θ=√3/2
2θ=60o,120o
(A)θ=30o
μ=u2sin2θ/2g=(√20)2sin230/2x10=0.25
θ=60°
μ=(√20)2sin230/2x10=3x0.25
(B) Umin=ucosθ=√20cos30°=√5m/s
(C) T=2usinθ/g For θ=30°
��=2x√20sin260°/10=√3/5
(D)Umax=mgh=1x10x0.25=2.5J
(√20)2sin2θ/10=√3
sin2θ=√3/2
2θ=60o,120o
(A)θ=30o
μ=u2sin2θ/2g=(√20)2sin230/2x10=0.25
θ=60°
μ=(√20)2sin230/2x10=3x0.25
(B) Umin=ucosθ=√20cos30°=√5m/s
(C) T=2usinθ/g For θ=30°
��=2x√20sin260°/10=√3/5
(D)Umax=mgh=1x10x0.25=2.5J
Two bodies of different masses ma and mb are dropped from two different heights, viz a and b. The ratio of times taken by the two to drop through these distances is- a) a : b
- b)
- c)
- d)a2 : b2
Correct answer is option 'C'. Can you explain this answer?
Two bodies of different masses ma and mb are dropped from two different heights, viz a and b. The ratio of times taken by the two to drop through these distances is
a)
a : b
b)
c)
d)
a2 : b2
|
|
Hansa Sharma answered |
Time taken by anybody to fall from a height h is
T = 2gh which is independent of the mass
Hence the required ratio will be that of option c
T = 2gh which is independent of the mass
Hence the required ratio will be that of option c
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 displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point 
- a)C
- b)D
- c)E
- d)F
Correct answer is option 'C'. Can you explain this answer?
The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point
a)
C
b)
D
c)
E
d)
F
|
|
Naina Sharma answered |
If we will draw a tangent of the point then ot will make angle x with t. Since only the shortest angle will be taken, thus it will also make 180-x.Since we now know that the slope= tan x, thus the slope at E = tan(180-x) =-tanx, thus the instantaneous velocity at E is negative
In the given question C and F have a positive slope so instantaneous velocity is positive. Since point D has zero slope so instantaneous velocity is zero but E has negative so instantaneous velocity is negative at E.
The figure shows a velocity-time graph of a particle moving along a straight line 
If the particle starts from the position x0 = -15 m, then its position at t = 2s will be- a) - 5m
- b)5m
- c)10m
- d)15m
Correct answer is option 'A'. Can you explain this answer?
The figure shows a velocity-time graph of a particle moving along a straight line
If the particle starts from the position x0 = -15 m, then its position at t = 2s will be
a)
- 5m
b)
5m
c)
10m
d)
15m
|
|
Riya Banerjee answered |
Displacement of the particle in 2 seconds is the area under v-t graph till 2 seconds,
s = (1/2)(10)(2) = 10 m
Net displacement = -15 + 10 = -5 m
s = (1/2)(10)(2) = 10 m
Net displacement = -15 + 10 = -5 m
Can you explain the answer of this question below:A hall has the dimensions 10m × 10m × 10 m. A fly starting at one corner ends up at a diagonally opposite corner. The magnitude of its displacement is nearly
- A:
- B:
- C:
- D:
The answer is B.
A hall has the dimensions 10m × 10m × 10 m. A fly starting at one corner ends up at a diagonally opposite corner. The magnitude of its displacement is nearly
|
|
Preeti Iyer answered |
Given: Dimensions of Hall = 10 m × 10 m × 10 m
To find: Displacement from one corner to another corner
Hall is in the shape of Cube.
Length of longest diagonal of cube is given by,
Diagonal of cube=√3 × Length of edge
Displacement = √3 × 10
Displacement=10√3
So The answer B is correct
To find: Displacement from one corner to another corner
Hall is in the shape of Cube.
Length of longest diagonal of cube is given by,
Diagonal of cube=√3 × Length of edge
Displacement = √3 × 10
Displacement=10√3
So The answer B is correct
The angle of projection of a body is 15º . The other angle for which the range is the same as the first one is equal to-- a) 30º
- b)45º
- c) 60º
- d)75º
Correct answer is option 'D'. Can you explain this answer?
The angle of projection of a body is 15º . The other angle for which the range is the same as the first one is equal to-
a)
30º
b)
45º
c)
60º
d)
75º
|
|
Neha Sharma answered |
We know that horizontal range R depends upon sin 2a where a is angle of projection. Also we know that sin 2a repeats its value for all π/2 -a. thus as here a = 15, the value of range will be equal to the same for 90 -15 = 75.
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
Two trains are each 50 m long moving parallel to each other at speeds 10 m/s and 15 m/s respectively, at what time will they pass each other? - a)8 s
- b)4 s
- c)2 s
- d)6 s
Correct answer is option 'B'. Can you explain this answer?
Two trains are each 50 m long moving parallel to each other at speeds 10 m/s and 15 m/s respectively, at what time will they pass each other?
a)
8 s
b)
4 s
c)
2 s
d)
6 s
|
|
Preeti Iyer answered |
As both the trains are moving in the opposite direction, the relative speed of each train with respect to each other be, v=10+15=25m/s
Here distance covered by each train = sum of their lengths =50+50=100m
∴ Required time=100/25=4sec
Here distance covered by each train = sum of their lengths =50+50=100m
∴ Required time=100/25=4sec
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.
Can you explain the answer of this question below:Two trains each of length 50 m are approaching each other on parallel rails. Their velocities are 10 m/sec and 15 m/sec. They will cross each other in -
- A:
2 Sec
- B:
4 Sec
- C:
10 Sec
- D:
6 Sec
The answer is a.
Two trains each of length 50 m are approaching each other on parallel rails. Their velocities are 10 m/sec and 15 m/sec. They will cross each other in -
2 Sec
4 Sec
10 Sec
6 Sec
|
|
Preeti Iyer answered |
Total distance =50+50=100
Total speed=10+15=25 s.
As we know, s= d/t [s = speed (meters/second)
d = distance traveled (meters)
t = time (seconds)]
we get t= d÷s
so, t=100/25
t= 4 sec
Total speed=10+15=25 s.
As we know, s= d/t [s = speed (meters/second)
d = distance traveled (meters)
t = time (seconds)]
we get t= d÷s
so, t=100/25
t= 4 sec
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
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
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.
Chapter doubts & questions for Kinematics of Motion in a Straight Line - Mathematics for A Level 2025 is part of A Level exam preparation. The chapters have been prepared according to the A Level exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for A Level 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Kinematics of Motion in a Straight Line - Mathematics for A Level in English & Hindi are available as part of A Level exam.
Download more important topics, notes, lectures and mock test series for A Level Exam by signing up for free.
Mathematics for A Level
124 videos|166 docs|207 tests
|
