Question

An object moving with the speed of 6.25 m/s is decelerated at a rate given by dv/dt = -2.5 √v where v is the instantaneous speed . the time taken by the object to come to rest , would be
a. 2s
b. 4s
c. 8s
d. 1s​
plz answer
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Answer 1

Answer:

Option A . 2 sec

Explanation:

Explanation is shown in the attached picture .

Thank you !!

Answer 2

TIME TAKEN IS 2 SECONDS (option a)

Given:

$a = \dfrac{dv}{dt} = - 2.5 \sqrt{v}$

To find:

Time at which velocity becomes zero ?

Calculation:

$\dfrac{dv}{dt} = - 2.5 \sqrt{v}$

$\implies\dfrac{dv}{ \sqrt{v} } = - 2.5t$

• Integrating on both sides with limits:

$\displaystyle\implies \int_{6.25}^{0}\dfrac{dv}{ \sqrt{v} } = - \int_{0}^{t} 2.5t$

$\displaystyle\implies \int_{0}^{6.25}\dfrac{dv}{ \sqrt{v} } = \int_{0}^{t} 2.5t$

$\displaystyle\implies 2 \sqrt{v} \bigg| _{0}^{6.25} = 2.5t - 0$

$\displaystyle\implies 2 \sqrt{6.25} = 2.5t$

$\displaystyle\implies 2 \times 2.5 = 2.5t$

$\displaystyle\implies t = 2 \: sec$

So, time taken is 2 seconds.