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Question: An object, moving with a speed of \(6.25\,m{s^{ - 1}}\) , is decelerated at a rate given by: \(\dfra...

An object, moving with a speed of 6.25ms16.25\,m{s^{ - 1}} , is decelerated at a rate given by: dvdt=2.5v\dfrac{{dv}}{{dt}} = - 2.5\sqrt v where vv is the instantaneous speed. The time taken by the object, to come to rest, would be:
A) 2s2s
B) 4s4s
C) 8s8s
D) 1s1s

Explanation

Solution

The deceleration rate is given in terms of the differential equation; we need to calculate the time taken for the body to come to rest. Integrate the equation to get the instantaneous speed in terms of time and then put the value of initial speed as 6.25ms16.25\,m{s^{ - 1}} to find the time taken.

Complete step by step solution:
We are given with the initial speed, u=6.25ms1u = 6.25\,m{s^{ - 1}}
The rate of deceleration is given as:
dvdt=2.5v\dfrac{{dv}}{{dt}} = - 2.5\sqrt v
The negative sign indicates that the body is decelerating.
dvv=2.5dt\Rightarrow \dfrac{{dv}}{{\sqrt v }} = - 2.5dt
Integrating the above equation, we get
6.250dvv=2.50tdt\Rightarrow \int\limits_{6.25}^0 {\dfrac{{dv}}{{\sqrt v }}} = - 2.5\int\limits_0^t {dt}
2v6.250=2.5t\therefore \left| {2\sqrt v } \right|_{6.25}^0 = - 2.5t
Let the time taken be tt seconds and as the body comes to rest therefore, the final velocity will be zero.
Substituting the limits, we get:
26.25=2.5t\Rightarrow 2\sqrt {6.25} = - 2.5t
t=2s\therefore t = 2s
Therefore, the time taken by the object to come to rest is 22 seconds.
Option A is the correct option.
Additional details: The acceleration of a body is the rate of change in velocity and it can be expressed as dvdt\dfrac{{dv}}{{dt}} . Similarly, as velocity is rate of change of displacement, it can also be represented as dxdt\dfrac{{dx}}{{dt}}. Deceleration is the negative of acceleration. For a body which is decelerating, it will eventually come to rest at some point in time. Instantaneous speed is the speed of an object at a particular moment in time. If we include the direction with that speed, we get the instantaneous velocity.

Note: Be careful with the limits after integrating. Remember that as the body comes to rest eventually hence, its final speed must be zero. Also remember that the negative sign indicates that the body is decelerating. As time cannot be negative hence, we have taken only positive values of time.