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All questions of Kinematic Equations for EmSAT Achieve Exam
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 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.
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
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
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.
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 
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
Can you explain the answer of this question below:An athlete uniformly increases his speed from rest at the rate of 1 m/s2. The time taken to complete 200 m race is:
- A:
10 s
- B:
14 s
- C:
5 s
- D:
20 s
The answer is d.
An athlete uniformly increases his speed from rest at the rate of 1 m/s2. The time taken to complete 200 m race is:
10 s
14 s
5 s
20 s
|
Ashish Saha answered |
v= 80*1000/3600 = 22.22 u=20*1000/3600=5.55 , Now a = v-u/t = 22.22-5.55/2 = 8.335
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
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
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.
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
Can you explain the answer of this question below:A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:- A:19 m
- B:17 m
- C:16 m
- D:18 m
The answer is a.
A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:
A:
19 m
B:
17 m
C:
16 m
D:
18 m
|
Sarita Reddy answered |
Given :
u = 36 km/h = 36x (5/18) = 10 m/s
Time of reaction, t = 0.9 s
decelration, a = -5 m/s^2
v = 0 m/s
Let s1 be the distance covered by man when the object is seen by him
we have, u = s1/t
because s1 = u x t
= 10 x 0.9
= 9 m
when he applies brakes and decelerates at the rate of 5 m/s^2, the distance covered by him is s2
We have, v^2 = u^2 + 2as2
therefore 0 = u^2 - 2as2 ( therefore v=0 m/s)
therefore s2 = u^2/2a = 10^2/ (2 x 5)
therefore s2 =10 m
So, total distance covered s = s1 + s2 = 9 +10 =19 m
Hence 19 m distance covered by man before he stops.
The displacement time graphs of two particles A and B are straight lines making angles of respectively 30º and 60º with the time axis. If the velocity of A is vA and that of B is vB then the value of 
is- a)1/2
- b)
- c)
- d)1/3
Correct answer is option 'D'. Can you explain this answer?
The displacement time graphs of two particles A and B are straight lines making angles of respectively 30º and 60º with the time axis. If the velocity of A is vA and that of B is vB then the value of is
isa)
1/2
b)
c)
d)
1/3
|
|
Suresh Reddy answered |
For some point of time say time t, the velocities of A and B would be va and vb respectively.
Hence we see that for both the graphs we get va/t = tan 30 and vb/t = tan 60
Thus we get the required ratio as tan 30 / tan 60 = ⅓
Hence we see that for both the graphs we get va/t = tan 30 and vb/t = tan 60
Thus we get the required ratio as tan 30 / tan 60 = ⅓
A body starts from rest and is uniformly accelerated for 30 s. The distance travelled in the first 10s is x1, next 10 s is x2 and the last 10 s is x3. Then x1 : x2 : x3 is the same as- a)1 : 2 : 4
- b)1 : 2 : 5
- c)1 : 3 : 5
- d) 1 : 3 : 9
Correct answer is option 'C'. Can you explain this answer?
A body starts from rest and is uniformly accelerated for 30 s. The distance travelled in the first 10s is x1, next 10 s is x2 and the last 10 s is x3. Then x1 : x2 : x3 is the same as
a)
1 : 2 : 4
b)
1 : 2 : 5
c)
1 : 3 : 5
d)
1 : 3 : 9
|
|
Preeti Iyer answered |
Let's say that the uniform acceleration is a.
Now distance travelled in 0 - 10 sec, x1 = ½.a x 100 = 50a
Similarly, distance travelled in 0 - 20 sec, = ½.a x 400 = 200a
Thus we get distance travelled in 10 - 20 sec, x2 = 200a - 50a = 150a
And, distance travelled in 0 - 30 sec, = ½.a x 900 = 450a
Thus we get distance travelled in 20 - 30 sec, x3 = 450a - 200a = 250a
Hence we get the ratio as 1 : 3 : 5
Now distance travelled in 0 - 10 sec, x1 = ½.a x 100 = 50a
Similarly, distance travelled in 0 - 20 sec, = ½.a x 400 = 200a
Thus we get distance travelled in 10 - 20 sec, x2 = 200a - 50a = 150a
And, distance travelled in 0 - 30 sec, = ½.a x 900 = 450a
Thus we get distance travelled in 20 - 30 sec, x3 = 450a - 200a = 250a
Hence we get the ratio as 1 : 3 : 5
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.
In the following velocity-time graph of a body, the distance and displacement travelled by the body in 5 second in meters will be - 
- a)75,115
- b) 105, 75
- c)45, 75
- d) 95, 55
Correct answer is option 'B'. Can you explain this answer?
In the following velocity-time graph of a body, the distance and displacement travelled by the body in 5 second in meters will be -
a)
75,115
b)
105, 75
c)
45, 75
d)
95, 55
|
EduRev JEE answered |
From the given velocity graph we get that displacement is w=equal to the area under the curve taken with sign while distance is area under the curve after the modulus is taken at first.
Thus displacement is
( ½ x 20 x 2 + 20 x 3 + ½ x 20 x 1) - (½ x 30 x 1) = 90 - 15 = 75m
And the distance will be
( ½ x 20 x 2 + 20 x 3 + ½ x 20 x 1) + (½ x 30 x 1) = 90 + 15 = 105m
Thus displacement is
( ½ x 20 x 2 + 20 x 3 + ½ x 20 x 1) - (½ x 30 x 1) = 90 - 15 = 75m
And the distance will be
( ½ x 20 x 2 + 20 x 3 + ½ x 20 x 1) + (½ x 30 x 1) = 90 + 15 = 105m
A particle, after starting from rest , experiences, constant acceleration for 20 seconds. If it covers a distance of S1, in first 10 seconds and distance S2 in next 10 sec, then- a)S2 = S1/2
- b) S2 = S1
- c)S2 = 2S1
- d)S2 = 3S1
Correct answer is option 'D'. Can you explain this answer?
A particle, after starting from rest , experiences, constant acceleration for 20 seconds. If it covers a distance of S1, in first 10 seconds and distance S2 in next 10 sec, then
a)
S2 = S1/2
b)
S2 = S1
c)
S2 = 2S1
d)
S2 = 3S1
|
|
Neha Joshi answered |
Let constant acceleration is a
distance travels in 20 seconds
s = 0*20+a*20²/2
s = 200a -------------(1)
distance travels in first 10 seconds
s1 = 0*10+a*10²/2
s1 = 50a -----------------(2)
distance travels in next 10 seconds
s2 = s-s1
s2 = 200a-50a = 150a ----------------(3)
s2/s1 = 150a/50a = 3
A body starts from rest, the ratio of distances travelled by the body during 3rd and 4thseconds is :- a)7/5
- b)5/7
- c)7/3
- d)3/7
Correct answer is option 'B'. Can you explain this answer?
A body starts from rest, the ratio of distances travelled by the body during 3rd and 4thseconds is :
a)
7/5
b)
5/7
c)
7/3
d)
3/7
|
Geetika Shah answered |
The velocity after 2 sec = 2a and velocity after 3 sec = 3a
For some constant acceleration a,
Now distance travelled in third second, s3 = 2a + ½ a
= 5/2 a
Similarly distance travelled in fourth second, s4= 3a + ½ a
= 7/2 a
Hence the reqiuired ratio is 5/7
For some constant acceleration a,
Now distance travelled in third second, s3 = 2a + ½ a
= 5/2 a
Similarly distance travelled in fourth second, s4= 3a + ½ a
= 7/2 a
Hence the reqiuired ratio is 5/7
A body is dropped from a height h under acceleration due to gravity g. If t1 and t2 are time intervals for its fall for first half and the second half distance, the relation between them is- a) t1 = t2
- b)t1 = 2t2
- c) t1 = 2.414 t2
- d)t1 = 4t2
Correct answer is option 'C'. Can you explain this answer?
A body is dropped from a height h under acceleration due to gravity g. If t1 and t2 are time intervals for its fall for first half and the second half distance, the relation between them is
a)
t1 = t2
b)
t1 = 2t2
c)
t1 = 2.414 t2
d)
t1 = 4t2
|
Jayant Mishra answered |
h=1/2g(t1+t2)2
2h/g=(t1+t2)2------(1)
h/2=1/2gt12
t12=h/g---------(2)
From 1st and 2nd
t12=(t1+t2)2/2
t1=(t1+t2)/1.41
1.41t1=t1+t2
t1=t2/.41
So
t1=2.414t2
At an instant t , the co-ordinates of a particle are x = at2, y = bt2 and z = 0 , then its velocity at the instant t will be- a) t
- b)2t
- c)
- d) 2t2
Correct answer is option 'B'. Can you explain this answer?
At an instant t , the co-ordinates of a particle are x = at2, y = bt2 and z = 0 , then its velocity at the instant t will be
a)
t
b)
2t
c)
d)
2t2
|
|
Neha Joshi answered |
At some time instant t, the displacement of the particle from origin = √(at2 )2+ (bt2)2
I.e. s = t2√a2+b2
Thus we get v = 2t√a2+b2
I.e. s = t2√a2+b2
Thus we get v = 2t√a2+b2
A man can swim in still water at a speed of 6 kmph and he has to cross the river and reach just opposite point on the other bank. If the river is flowing at a speed of 3 kmph, and the width of the river is 2 km, the time taken to cross the river is (in hours) - a) 2/27
- b) 2/√27
- c) 2/3
- d) 2/√45
Correct answer is option 'B'. Can you explain this answer?
A man can swim in still water at a speed of 6 kmph and he has to cross the river and reach just opposite point on the other bank. If the river is flowing at a speed of 3 kmph, and the width of the river is 2 km, the time taken to cross the river is (in hours)
a)
2/27
b)
2/√27
c)
2/3
d)
2/√45
|
Ananya Das answered |
To reach the exact opposite end the parallel component of velocity would nullify the velocity of the river, hence we get 6.sin a = 3 , where a is the angle with perpendicular to river flow.
Thus we get sin a = ½ and cos a =√3/2
Thus the perpendicular component of velocity, 6.cos a = 3√3
Thus time taken is 2 / 3√3
Thus we get sin a = ½ and cos a =√3/2
Thus the perpendicular component of velocity, 6.cos a = 3√3
Thus time taken is 2 / 3√3
Vectors can be added by- a)adding the magnitudes of the vectors
- b)adding the angles of the vectors
- c)translating the two vectors
- d)parallelogram law of addition
Correct answer is option 'D'. Can you explain this answer?
Vectors can be added by
a)
adding the magnitudes of the vectors
b)
adding the angles of the vectors
c)
translating the two vectors
d)
parallelogram law of addition
|
Puja Kaur answered |
Explanation:Parallelogram law of vector addition is,If two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is a diagonal passing through the point of contact of two vectors.
The equations of motion can be derived by using:- a)Distance – time graph
- b)Velocity – time graph for non-uniform acceleration
- c)Displacement time graph
- d)Velocity – time graph for uniform acceleration
Correct answer is 'D'. Can you explain this answer?
The equations of motion can be derived by using:
a)
Distance – time graph
b)
Velocity – time graph for non-uniform acceleration
c)
Displacement time graph
d)
Velocity – time graph for uniform acceleration
|
Indu Gupta answered |
Consider a velocity - time graph for a uniformly accelerated body starting from rest is represented as follows.
u = velocity at time t1
v = velocity at time t2
If acceleration is represented as a, then, acceleration is defined as the rate of change in velocity.
⇒ a = v-u / t2 - t1
⇒ a = v-u /t
Or,v-u = at
⇒ v= u + at
u = velocity at time t1
v = velocity at time t2
If acceleration is represented as a, then, acceleration is defined as the rate of change in velocity.
⇒ a = v-u / t2 - t1
⇒ a = v-u /t
Or,v-u = at
⇒ v= u + at
A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:- a)19 m
- b)17 m
- c)16 m
- d)18 m
Correct answer is 'A'. Can you explain this answer?
A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:
a)
19 m
b)
17 m
c)
16 m
d)
18 m
|
|
Arvind Singh answered |
Given . u = 36 km/h=36x (5/18)=10 m/s Time of reaction, t=0.9 s decelration, a=-5 m/s^2 v = 0 m/s Let s1 be the distance covered by man when the object is seen by him we have, u = s1/t because s1 = u x t=10x0.9 =9 m when he applies brakes and decelerates at the rate of 5 m/s^2, the distance covered by him is s2 We have, v^2 = u^2 + 2as2 because 0 = u^2 - 2as2 (because v=0 m/s) because s2 = u^2/2a = 10^2/2x5 s2 =10 m So, total distance covered s = s1 + s2 = 9 + 10 =19 m Hence 19 m distance covered by man before he stops.
A car travels a distance of 2000 m. If the first half distance is covered at 40 km/hour and the second half at velocity v and if the average velocity is 48 km/hour, then the value of v is -- a) 56 km/hour
- b)60 km/hour
- c)50 km/hour
- d)48 km/hour
Correct answer is option 'B'. Can you explain this answer?
A car travels a distance of 2000 m. If the first half distance is covered at 40 km/hour and the second half at velocity v and if the average velocity is 48 km/hour, then the value of v is -
a)
56 km/hour
b)
60 km/hour
c)
50 km/hour
d)
48 km/hour
|
|
Neha Joshi answered |
Total distance covered is 2km, and first 1km is covered with speed of 40kmph thus total time taken is 1/40 hrs similarly as second half is travelled with velocity v time taken would be 1/v hrs. But we know that as average velocity is 48 kmph total time travelled can also be written as 2/48 hrs. Thus we get
1/40 + 1/v = 2/48
I.e 1/v = 1/24 - 1/40
= (40 - 24) / 40 x 24
Thus we get v = 40 x 24 / 16
= 60 kmph
1/40 + 1/v = 2/48
I.e 1/v = 1/24 - 1/40
= (40 - 24) / 40 x 24
Thus we get v = 40 x 24 / 16
= 60 kmph
The a-t graph of the particle is correctly shown by
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The a-t graph of the particle is correctly shown by
a)
b)
c)
d)
|
|
Riya Banerjee answered |
The acceleration of the particle moving in straight line is given as, a=d2x/dt2
Geometrically it is the concavity of the x−t graph. If the graph is concave upward then the double derivative is positive. If the graph is concave downwards then the double derivative is negative.Consider the provided x−t graph,
Since the x−t graph is a parabola which of second degree the a−t graph must be a constant line which is of zero degree. Since the parabola is concave downwards the acceleration must be negative.
The only graph which satisfies the above conditions is graph D.
Hence, option D is correct.
Geometrically it is the concavity of the x−t graph. If the graph is concave upward then the double derivative is positive. If the graph is concave downwards then the double derivative is negative.Consider the provided x−t graph,
Since the x−t graph is a parabola which of second degree the a−t graph must be a constant line which is of zero degree. Since the parabola is concave downwards the acceleration must be negative.
The only graph which satisfies the above conditions is graph D.
Hence, option D is correct.
A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is - a) 4v2/5g
- b)4g/5v2
- c)4v3/5g2
- d)4v/5g2
Correct answer is option 'A'. Can you explain this answer?
A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is
a)
4v2/5g
b)
4g/5v2
c)
4v3/5g2
d)
4v/5g2
|
|
Anjana Sharma answered |
Range => R = V^2sin2thetha/g
and H = V^2/2g
and given as the relation,
R = 2H
V^2sin2thetha/g = V^2/g
sin2thetha = 1
2thetha = pi/2
or
thetha = pi/4
Therfore the Range becomes =>
R = V^2/g
An observer moves with a constant speed along the line joining two stationary objects. He will observe that the two objects- a)Have the same speed
- b)Have the same velocity
- c)Move in the same direction
- d) Move in opposite directions
Correct answer is option 'A,B,C'. Can you explain this answer?
An observer moves with a constant speed along the line joining two stationary objects. He will observe that the two objects
a)
Have the same speed
b)
Have the same velocity
c)
Move in the same direction
d)
Move in opposite directions
|
|
Suresh Reddy answered |
Given constant speed this question so speed is same, since it moved on a straight line so, velocity is same and in the same direction.
The figure shows a velocity-time graph of a particle moving along a straight line The correct displacement-time graph of the particle is shown as- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The figure shows a velocity-time graph of a particle moving along a straight line
The correct displacement-time graph of the particle is shown as
a)
b)
c)
d)
|
Manish Aggarwal answered |
Displacement is said to the shortest path covered by a body or object. Hence, when the velocity is positive the object moves in one direction and when its negative then in the opposite direction thus making the displacement first increase then decrease.
Which motion does not require force to maintain it? - a)Uniform circular motion
- b)Elliptical motion
- c)Uniform straight line motion
- d)Projectile motion
Correct answer is option 'C'. Can you explain this answer?
Which motion does not require force to maintain it?
a)
Uniform circular motion
b)
Elliptical motion
c)
Uniform straight line motion
d)
Projectile motion
|
|
Nandini Iyer answered |
Uniform straight line motion is a type of linear motion. It works according to Newton's First law of motion i.e. objects that do not experience any net force will continue to move in a straight line with a constant velocity until they are subjected to a net force.
An object A is moving with 10 m/s and B is moving with 5 m/s in the same direction of positive x-axis. A is 100 m behind B as shown. Find time taken by A to Meet B 
- a)18 sec.
- b)16 sec.
- c) 20 sec.
- d)17 sec.
Correct answer is option 'C'. Can you explain this answer?
An object A is moving with 10 m/s and B is moving with 5 m/s in the same direction of positive x-axis. A is 100 m behind B as shown. Find time taken by A to Meet B
a)
18 sec.
b)
16 sec.
c)
20 sec.
d)
17 sec.
|
Infinity Academy answered |
vA=10m/s
vB=5m/s
Here, Vrelative=10m/s-5m/s
=5m/s
Now,t=S/V=100m/5m/s=20sec
vB=5m/s
Here, Vrelative=10m/s-5m/s
=5m/s
Now,t=S/V=100m/5m/s=20sec
A particle covers half of the circle of radius r. Then the displacement and distance of the particle are respectively -- a)2πr, 0
- b) 2r, πr
- c)
, 2r - d) πr, r
Correct answer is option 'B'. Can you explain this answer?
A particle covers half of the circle of radius r. Then the displacement and distance of the particle are respectively -
a)
2πr, 0
b)
2r, πr
c)
, 2rd)
πr, r
|
|
Naina Bansal answered |
This solely depends upon the path particle is adopting.
1 If covering half the circlemeans starting from any point on circumference and reaching the center,then
Distance = Displacement=r
2 But if covering half the circle means moving on cicumference and covering half circle (which I think case is),then
Distance= (pie)*r Dispacement=2r
A body performs an accelerated motion, with uniform speed. The motion of body is- a)Linear
- b)Circular
- c)Parabolic
- d)Irregular
Correct answer is 'B'. Can you explain this answer?
A body performs an accelerated motion, with uniform speed. The motion of body is
a)
Linear
b)
Circular
c)
Parabolic
d)
Irregular
|
|
Shounak Chauhan answered |
Explanation:
When a body performs an accelerated motion with uniform speed, its motion is circular. The reason behind it can be explained in the following points:
Uniform Speed:
Uniform speed means the body is moving with a constant speed. It means that the magnitude of the velocity of the body is constant. The direction of the velocity vector can be changed, but the magnitude of the velocity remains constant.
Accelerated Motion:
Accelerated motion means the velocity of the body is changing with time. It means that the magnitude of the acceleration of the body is not zero. The direction of the acceleration vector can be changed, but the magnitude of the acceleration is not zero.
Circular Motion:
Circular motion is a kind of motion in which a body moves in a circular path. In circular motion, the direction of the velocity vector changes continuously, but the magnitude of the velocity remains constant. The direction of the acceleration vector also changes continuously, but the magnitude of the acceleration is not zero.
Conclusion:
When a body performs an accelerated motion with uniform speed, its motion is circular. In this case, the magnitude of the velocity of the body remains constant, but the direction of the velocity vector changes continuously. The magnitude of the acceleration of the body is not zero, and the direction of the acceleration vector also changes continuously. Hence, the motion of the body is circular.
When a body performs an accelerated motion with uniform speed, its motion is circular. The reason behind it can be explained in the following points:
Uniform Speed:
Uniform speed means the body is moving with a constant speed. It means that the magnitude of the velocity of the body is constant. The direction of the velocity vector can be changed, but the magnitude of the velocity remains constant.
Accelerated Motion:
Accelerated motion means the velocity of the body is changing with time. It means that the magnitude of the acceleration of the body is not zero. The direction of the acceleration vector can be changed, but the magnitude of the acceleration is not zero.
Circular Motion:
Circular motion is a kind of motion in which a body moves in a circular path. In circular motion, the direction of the velocity vector changes continuously, but the magnitude of the velocity remains constant. The direction of the acceleration vector also changes continuously, but the magnitude of the acceleration is not zero.
Conclusion:
When a body performs an accelerated motion with uniform speed, its motion is circular. In this case, the magnitude of the velocity of the body remains constant, but the direction of the velocity vector changes continuously. The magnitude of the acceleration of the body is not zero, and the direction of the acceleration vector also changes continuously. Hence, the motion of the body is circular.
A body starts to slide over a horizontal surface with an initial velocity of 0.2 m/s. Due to friction, its velocity decreases at the rate of 0.02 m/s2. How much time will it take for the body to stop?- a)10 s
- b)15s
- c)1s
- d)5s
Correct answer is option 'A'. Can you explain this answer?
A body starts to slide over a horizontal surface with an initial velocity of 0.2 m/s. Due to friction, its velocity decreases at the rate of 0.02 m/s2. How much time will it take for the body to stop?
a)
10 s
b)
15s
c)
1s
d)
5s
|
|
Anita Menon answered |
Initial velocity, u = 0.2 m/s.
Final velocity, v = 0
Acceleration, a = 0.02 m/s^2.
v = u + at
0 = 0.2 - 0.02 x t
0.02 x t = 0.2
t = 0.2/0.02 = 10 sec.
Thus, the body will take 10 seconds to stop.
Let v and a denote the velocity and acceleration respectively of a body
a) A can be non zero when v = 0
b) A must be zero when v = 0
c) A may be zero when v ≠ 0
d) The direction of a must have some correlation with the direction of v.
Correct answer is option 'A,C'. Can you explain this answer?
|
|
Neha Joshi answered |
A can be zero when velocity and vice-versa you can get that when you see a pendulum at mean a = 0 but v = max and at extreme a = max when v = 0. Also a “can” be zero when v is zero.
The figure shows a velocity-time graph of a particle moving along a straight line Choose the incorrect statement. The particle comes to rest at- a)t = 0 s
- b)t = 5 s
- c)t = 8 s
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
The figure shows a velocity-time graph of a particle moving along a straight line
Choose the incorrect statement. The particle comes to rest at
a)
t = 0 s
b)
t = 5 s
c)
t = 8 s
d)
None of these
|
|
Riya Banerjee answered |
The particle comes to rest (v = 0) at two instants. It comes to rest first time at somewhere between 4th second and 6th second. At t = 8 second it again comes to rest.
We will find when the particle comes to rest for the first time. Between 4th and 6th second (time interval of 2 seconds) the particle changes it's velocity from 10 m/s to -20 m/s. So, the acceleration on the particle during this part of motion is,
a = (-20 - 10)/2 = -15 seconds
The particle comes to rest when v = 0
v - u = at
0 - 10 = (-15)t
⇒ t = 1.5 seconds
So the particle come to rest at (4 + 1.5)th = 5.5 th second and 8th second
We will find when the particle comes to rest for the first time. Between 4th and 6th second (time interval of 2 seconds) the particle changes it's velocity from 10 m/s to -20 m/s. So, the acceleration on the particle during this part of motion is,
a = (-20 - 10)/2 = -15 seconds
The particle comes to rest when v = 0
v - u = at
0 - 10 = (-15)t
⇒ t = 1.5 seconds
So the particle come to rest at (4 + 1.5)th = 5.5 th second and 8th second
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