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:

(1) 5: 9

(2) 5: 7

(3) 9: 5

(4) 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  

(1) $10\times {10}^4$ m/s²

(2) $12\times {10}^4$ m/s²

(3) $13.5\times {10}^4$ m/s²

(4) $15\times {10}^4$ 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 \(50\) metres from the bus. As soon as the bus begins its motion with an acceleration of \(1\) ms^–2, the student starts running towards the bus with a uniform velocity \(u\). Assuming the motion to be along a straight road, the minimum value of \(u\), so that the student is able to catch the bus is:
1. \(5\) ms^–1
2. \(8\) ms^–1
3. \(10\) ms^–1
4. \(12\) 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. $\frac{2v}{a}$

2. $\frac{v}{a}$

3. $\frac{v}{2a}$

4. $\sqrt{\frac{v}{2a}}$ 

A particle moves along X-axis in such a way that its coordinate X varies with time t according to the equation $x=(2-5t+6t^2)m$. 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) $\frac{aT}{4}$

(2) $\frac{3aT}{2}$

(3) $\frac{aT}{2}$

(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