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. \(128~\text{m}\)

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

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. $15m/s^2$

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  

1. 50

2. 5

3. 2

4. 0.5

If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s. Then it covers a distance of 

1. 20 m

2. 400 m

3. 1440 m

4. 2880 m

If a train travelling at 72 kmph is to be brought to rest in a distance of 200 metres, then its retardation should be

1. 20 ms^–2

2. 10 ms^–2

3. 2 ms^–2

4.1 ms^–2

The displacement of a particle starting from rest (at t = 0) is given by $s=6t^2-t^3$. The time in seconds at which the particle will attain zero velocity again, is  

1. 2

2. 4

3. 6

4. 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?

1. 10 sec

2. 20 sec

3. 30 sec

4. 35 sec

The motion of a particle is described by the equation $x=a+b{t}^2$ where a = 15 cm and b = 3 cm/s². Its instantaneous velocity at time 3 sec will be 

1. 36 cm/sec

2. 18 cm/sec

3. 16 cm/sec

4. 32 cm/sec