If body having initial velocity zero is moving with uniform acceleration 8 m/sec² , then the distance travelled by it in fifth second will be  

1. 36 metres

2. 40 metres

3. 100 metres

4. Zero

An alpha particle enters a hollow tube of 4 m length with an initial speed of 1 km/s. It is accelerated in the tube and comes out of it with a speed of 9 km/s. The time for which it remains inside the tube is

1. $8\times {10}^{-3} s$

2. $80\times {10}^{-3}s$

3. $800\times {10}^{-3}s$

4. $8\times {10}^{-4}s$

Two cars A and B are travelling in the same direction with velocities v₁ and v₂ $(v_1>v_2)$. When the car A is at a distance d behind car B, the driver of the car A applied the brake producing uniform retardation a. There will be no collision when-

1. $d<\frac{{(v_1-v_2)}^2}{2a}$

2. $d<\frac{v_1^2-v_2^2}{2a}$

3. $d>\frac{{(v_1-v_2)}^2}{2a}$

4. $d>\frac{v_1^2-v_2^2}{2a}$ 

A body of mass 10 kg is moving with a constant velocity of 10 m/s. When a constant force acts for 4 seconds on it, it moves with a velocity 2 m/sec in the opposite direction. The acceleration produced in it is 

1. 3 m/sec²

2. –3 m/sec²

3. 0.3 m/sec²

4. –0.3 m/sec²

A body starts from rest from the origin with an acceleration of \(6~\text{m/s}^2\) along the \(x\text-\)axis and \(8~\text{m/s}^2\) along the \(y\text-\)axis. Its distance from the origin after \(4\) seconds will be:
1. \(56~\text{m}\)
2. \(64~\text{m}\)
3. \(80~\text{m}\)
4. \(128~\text{m}\)

A car moving with a velocity of 10 m/s can be stopped by the application of a constant force F in a distance of 20 m. If the velocity of the car is 30 m/s, it can be stopped by this force in 

1. $\frac{20}{3}m$

2. 20 m

3. 60 m

4. 180 m

The displacement of a particle is given by \(y = a + bt + ct^{2} - dt^{4}\). The initial velocity and acceleration are, respectively:

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 ?

1. 8 m

2. 2 m

3. 4 m

4. 6 m

An elevator car, whose floor to ceiling distance is equal to \(2.7~\text{m}\), starts ascending with constant acceleration of \(1.2~\text{ms}^{-2}\). \(2\) sec after the start, a bolt begins falling from the ceiling of the car. The free fall time of the bolt is: 
1. \(\sqrt{0.54}~\text{s}\)
2. \(\sqrt{6}~\text{s}\)
3. \(0.7~\text{s}\)
4. \(1~\text{s}\)

The displacement is given by $x=2t^2+t+5$, the acceleration at $t=2s$ is 

1. $4m/s^2$

2. $8m/s^2$

3. $10m/s^2$

4. $15m/s^2$