Q. No.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
1. 10×10410×104 m/s2
2. 12×10412×104 m/s2
3. 13.5×10413.5×104 m/s2
4. 15×10415×104 m/s2
A particle starts from rest, accelerates at 2 m/s2 for 10s and then goes for constant speed for 30s and then decelerates at 4 m/s2 till it stops. What is the distance travelled by it?
1. 750 m
2. 800 m
3. 700 m
4. 850 m
The engine of a motorcycle can produce a maximum acceleration 5 m/s2. Its brakes can produce a maximum retardation 10 m/s2. What is the minimum time in which it can cover a distance of 1.5 km?
1. 30 sec
2. 15 sec
3. 10 sec
4. 5 sec
A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is
1. 6m
2. 12m
3. 18m
4. 24m
A student is standing at a distance of 5050 metres from the bus. As soon as the bus begins its motion with an acceleration of 11 ms–2, the student starts running towards the bus with a uniform velocity uu. Assuming the motion to be along a straight road, the minimum value of uu, so that the student is able to catch the bus is:
1. 55 ms–1
2. 88 ms–1
3. 1010 ms–1
4. 1212 ms–1
A body A moves with a uniform acceleration a and zero initial velocity. Another body B, starts from the same point moves in the same direction with a constant velocity v. The two bodies meet after a time t. The value of t is
1. 2va2va
2. vava
3. v2av2a
4. √v2a√v2a
A particle moves along X-axis in such a way that its coordinate X varies with time t according to the equation x=(2−5t+6t2) mx=(2−5t+6t2) m. The initial velocity of the particle is
1. –5 m/s
2. 6 m/s
3. –3 m/s
4. 3 m/s
A car starts from rest and moves with uniform acceleration 'a' on a straight road from time t = 0 to t = T. After that, a constant deceleration brings it to rest. In this process the average speed of the car is:
1. aT4aT4
2. 3aT23aT2
3. aT2aT2
4. aT
An object accelerates from rest to a velocity of 27.5 m/s in 10 sec . Then find the distance covered by the object in the next 10 sec:
1. 550 m
2. 137.5 m
3. 412.5 m
4. 275 m
If the velocity of a particle is given by v=(180−16x)1/2 m/sv=(180−16x)1/2 m/s, then its acceleration will be:
| 1. | zero | 2. | 8 m/s2 |
| 3. | −8 m/s2 | 4. | 4 m/s2 |
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