An object, moving with a speed of $6.25\, m/s$, is decelerated at a rate given by

$\frac{{dv}}{{dt}} =  - 2.5\sqrt v$

where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be ........$s$

Read each statement below carefully and state with reasons and examples, if it is true or false 

A particle in one-dimensional motion

$(a)$ with zero speed at an instant may have non-zero acceleration at that instant

$(b)$ with zero speed may have non-zero velocity.

$(c)$ with constant speed must have zero acceleration.

$(d)$ with positive value of acceleration must be speeding up.

The velocity $(v)-$ time $(t)$ plot of the motion of a body is shown below:

(image)

The acceleration $(a)-$ time $(t)$ graph that best suits this motion is :

Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity, or deceleration, $-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0 $ and $a$.

A particle starts moving along a line from zero initial velocity and comes to rest after moving distance $d$. During its motion, it had a constant acceleration $f$ over $2 / 3$ of the distance and covered the rest of the distance with constant retardation. The time taken to cover the distance is

The area under acceleration-time graph gives