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Question: An object moving with speed of \(6.25\text{ m/sec}\) , is de-accelerated at a rate given by\(\dfrac{...

An object moving with speed of 6.25 m/sec6.25\text{ m/sec} , is de-accelerated at a rate given bydvdt=−2.5v\dfrac{\text{dv}}{\text{dt}}=-2.5\sqrt{\text{v}} where v= instantaneous speed. What would be the time taken by the object, to come to rest?

Explanation

Solution

In the question , the rate of deceleration is given by the equation
dvdt=−2.5v\dfrac{\text{dv}}{\text{dt}}=-2. 5\sqrt{\text{v}}
We have to calculate the time taken by the object to stop or come to rest . So , on integrating this equation with respect to t and then, by applying proper limits, the value of t is calculated.

Complete answer:
Given: dvdt=−2.5v\dfrac{\text{dv}}{\text{dt}}=-2.5\sqrt{\text{v}}
On rearranging, we get
dvv=−2.5 dt\dfrac{\text{dv}}{\sqrt{\text{v}}}=-2.5\text{ dt}
Now limits of t will be from 0 to t and limits of v will be from v =6.256.25(at t = 0) to 0 (at time= t).
So, on integration, we get
∫6.25odvv=−2.5 ∫otdt\int\limits_{6.25}^{\text{o}}{\dfrac{\text{dv}}{\sqrt{\text{v}}}=-2.5}\text{ }\int\limits_{\text{o}}^{\text{t}}{\text{dt}}
∣v−1/2+1(−1/2+1)∣06.25 ==−2.5∣t∣t0{{\left| \dfrac{{{\text{v}}^{-1/2+1}}}{\left( -1/2+1 \right)} \right|}^{0}}_{6. 25}\text{ =}=-2. 5{{\left| \text{t} \right|}^{\text{t}}}_{0}
∣2v∣06.25 =−2.5 ∣t∣t0{{\left| 2\sqrt{\text{v}} \right|}^{0}}_{6. 25}\text{ }=-2.5\text{ }{{\left| \text{t} \right|}^{\text{t}}}_{0}
2(0−6.25)=−2.5 t −2×6.25=−2.5 t 2×2.5 =2.5 t t=2×2.5 2.5  t=2sec⁡ \begin{aligned} & 2\left( 0-\sqrt{6.25} \right)=-2. 5\text{ t} \\\ & -2\times \sqrt{6.25}=-2.5\text{ t} \\\ & 2\times 2.5\text{ =}2.5\text{ t} \\\ & \text{t}=\dfrac{2\times 2.5\text{ }}{2.5\text{ }} \\\ & \text{t}=2\sec \\\ \end{aligned}

So the correct option is (B).

Note:
Acceleration is defined strictly as the time rate of change of velocity vector. Deceleration, on the other hand, is acceleration that causes reduction in speed.
There is a difference between deceleration and negative acceleration.
Deceleration refers to the acceleration in the direction opposite to the direction of the velocity. Deceleration always reduces speed. Negative acceleration, however, is the acceleration in the negative direction in the chosen coordinate system. These two may or may not be the same.
When the motion is taken to be in one dimension, then the deceleration occurs when signs of velocity and acceleration are opposite. When there is a negative velocity and a positive acceleration, this means that there is deceleration and if there is a positive velocity and a negative acceleration , it means deceleration. A positive velocity and a positive acceleration means acceleration and a negative velocity and a negative acceleration means acceleration. This difference needs to be understood properly.
So, we summarize the discussion as
Deceleration results in a decrease in magnitude of velocity.
In one dimensional motion, the deceleration is defined as the acceleration which is opposite to the velocity.