A car moving with a speed of 40 km/h can be stopped by applying brakes for atleast 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance ?
8 m
2 m
4 m
6 m
An elevator car, whose floor to ceiling distance is equal to 2.7 m, starts ascending with constant acceleration of 1.2 ms–2. 2 sec after the start, a bolt begins falling from the ceiling of the car. The free fall time of the bolt is
0.54 s
6 s
0.7 s
1 s
The displacement is given by x=2t2+t+5, the acceleration at t=2s is
4 m/s2
8 m/s2
10 m/s2
15 m/s2
Two trains travelling on the same track are approaching each other with equal speeds of 40 m/s. The drivers of the trains begin to decelerate simultaneously when they are just 2.0 km apart. Assuming the decelerations to be uniform and equal, the value of the deceleration to barely avoid collision should be
11.8 m/s2
11.0 m/s2
1.6 m/s2
0.8 m/s2
A body moves from rest with a constant acceleration of 5 m/s2. Its instantaneous speed (in m/s) at the end of 10 sec is
50
5
2
0.5
A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation
Both will be equal
First will be half of second
First will be 1/4 of second
No definite ratio
A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd second
75
57
73
37
The acceleration ‘a’ in m/s2 of a particle is given by a=3t2+2t+2 where t is the time. If the particle starts out with a velocity, u = 2 m/s at t = 0, then the velocity at the end of 2 seconds will be
12 m/s
18 m/s
27 m/s
36 m/s
A particle moves along a straight line such that its displacement at any time t is given by S=t3−6t2+3t+4 metres. The velocity when the acceleration is zero is:
4 ms-1
−12 ms−1
42 ms−1
−9 ms−1
If a body starts from rest and travels 120 cm in the 6th second, then what is the acceleration
0.20 m/s2
0.027 m/s2
0.218 m/s2
0.03 m/s2
If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s. Then it covers a distance of
20 m
400 m
1440 m
2880 m
The position x of a particle varies with time t as x=at2−bt3. The acceleration of the particle will be zero at time t equal to
ab
2a3b
a3b
Zero
If a train travelling at 72 kmph is to be brought to rest in a distance of 200 metres, then its retardation should be
20 ms–2
10 ms–2
2 ms–2
1 ms–2
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
4
6
8
Two cars A and B are at rest at the same point initially. If A starts with uniform velocity of 40 m/sec and B starts in the same direction with a constant acceleration of 4 m/s2, then B will catch A after how much time?
10 sec
20 sec
30 sec
35 sec
The motion of a particle is described by the equation x=a+bt2 where a = 15 cm and b = 3 cm/s2. Its instantaneous velocity at time 3 sec will be
36 cm/sec
18 cm/sec
16 cm/sec
32 cm/sec
A body travels for 15 sec starting from rest with constant acceleration. If it travels distances S1, S2 and S3 in the first five seconds, second five seconds and next five seconds respectively the relation between S1, S2 and S3 is
S1=S2=S3
5S1=3S2=S3
S1=13S2=15S3
S1=15S2=13S3
A body is moving according to the equation x=at+bt2−ct3 where x = displacement and a, b and c are constants. The acceleration of the body is
a+2bt
2b+6ct
2b−6ct
3b−6ct2
A particle travels 10 m in first 5 sec and 10m in the next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec ?
8.3 m
9.3 m
10.3 m
None of above
The distance travelled by a particle is proportional to the squares of time, then the particle travels with
Uniform acceleration
Uniform velocity
Increasing acceleration
Decreasing velocity
Velocity of a particle changes when
Direction of velocity changes
Magnitude of velocity changes
Both of above
None of the above
The motion of a particle is described by the equation u = at, where u is velocity and a is a constant. The distance travelled by the particle in the first 4 seconds
4 a
12 a
6 a
8 a
The relation 3t=3x+6 describes the displacement of a particle in one direction where x is in metres and t in sec. The displacement, when velocity is zero, is
24 metres
12 metres
5 metres
The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 ms–1. If the change in velocity of the body is 0.18ms–1 during this time, its uniform acceleration is
0.01 ms–2
0.02 ms–2
0.03 ms–2
0.04 ms–2
Equation of displacement for any particle is s=3t3+7t2+14t+8m. Its acceleration at time t = 1 sec is
16 m/s2
25 m/s2
32 m/s2
The position of a particle moving along the x-axis at certain times is given below :
Which of the following describes the motion correctly?
Uniform, accelerated
Uniform, decelerated
Non-uniform, accelerated
There is not enough data for generalization
Consider the acceleration, velocity and displacement of a tennis ball as it falls to the ground and bounces back. Directions of which of these changes in the process ?
Velocity only
Displacement and velocity
Acceleration, velocity and displacement
Displacement and acceleration
The displacement of a particle, moving in a straight line, is given by s=2t2+2t+4 where s is in metres and t in seconds. The acceleration of the particle is
2 m/s2
6 m/s2
A body A starts from rest with an acceleration a1. After 2 seconds, another body B starts from rest with an acceleration a2. If they travel equal distances in the 5th second, after the start of A, then the ratio a1: a2 is equal to:
5: 9
5: 7
9: 5
9: 7
The velocity of a bullet is reduced from 200m/s to 100m/s while travelling through a wooden block of thickness 10cm. The retardation, assuming it to be uniform, will be
10×104 m/s2
12×104 m/s2
13.5×104 m/s2
15×104 m/s2
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