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) $15m/s^2$