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

If the velocity of a particle is given by \(v = (180-16x)^{1/2}~\text{m/s}\), then its acceleration will be: 

The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depends on time obtained 

1. $a\propto t^2$

2. $a\propto t^4$

3. $a\propto t^3$

4. $a\propto t$