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) $a\propto t$ 

Starting from rest, acceleration of a particle is $a=2(t-1).$ The velocity of the particle at $t=5s$ is:

1.  15 m/sec
2.  25 m/sec
3.  5 m/sec
4.  None of these

Speed of two identical cars are u and 4u at a specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is:

(1) 1 : 1

(2) 1 : 4

(3) 1 : 8

(4) 1 : 16

A body is moving with uniform acceleration describes 40 m in the first 5 sec and 65 m in the next 5 sec. Its initial velocity will be:

(1) 4 m/s

(2) 2.5 m/s

(3) 5.5 m/s

(4) 11 m/s 

The displacement x of a particle varies with time $t, x=a{e}^{-\alpha t}+b{e}^{\beta t}$, where $a,b,\alpha \text{\hspace{0.17em}and }\beta$ are positive constants. The velocity of the particle will 

1. Go on decreasing with time
2. Be independent of $\alpha$ and $\beta$
3. Drop to zero when $\alpha$ = $\beta$
4. Go on increasing with time

A car, starting from rest, accelerates at the rate f through a distance S, then continues at a constant speed for time t and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is 15 S, then, 

(1) $S=\frac{1}{2}f{t}^2$

(2) $S=\frac{1}{4}f{t}^2$

(3) $S=\frac{1}{72}f{t}^2$  

(4) $S=\frac{1}{6}f{t}^2$