An object- moving with a speed of 6-25 m-s- is decelerated a rate given by-frac-dv-dt- - -2-5sqrt-v- where is the instantaneous speed- The time taken by the object- to come to rest- would be-2 s4 s1 s8 s

Question

Toppr Admin · Accepted Answer

-xA0-Accordingtoquestion-x222B-06-25-dv-x221A-v-x2212-2-5-x222B-t0dt-x21D2-x2223-x2223-2-x221A-v-x2223-x2223-06-25-x2212-2-5t-x21D2-2-x221A-6-25-2-5t-x2234-t-2ssothecorrectoptuonisA

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:
$\frac{d v}{d t} = - 2.5 \sqrt{v}$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be:
2 s
4 s
1 s
8 s

$A c c o r d i n g t o q u e s t i o n . . . . . . . . . . . . . . . . . . . \int_{6.25}^{0} (\frac{d v}{\sqrt{v}}) = - 2.5 \int_{0}^{t} d t \Rightarrow {∣ ∣ 2 \sqrt{v} ∣ ∣}_{6.25}^{0} = - 2.5 t \Rightarrow 2. \sqrt{6.25} = 2.5 t ∴ t = 2 s s o t h e c o r r e c t o p t u o n i s A$

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:dvdt=−2.5√v where v is the instantaneous speed. The time taken by the object, to come to rest, would be:2 s4 s1 s8 s

Accordingtoquestion...................∫06.25(dv√v)=−2.5∫t0dt⇒∣∣2√v∣∣06.25=−2.5t⇒2.√6.25=2.5t∴t=2ssothecorrectoptuonisA

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:
$\frac{d v}{d t} = - 2.5 \sqrt{v}$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be:
2 s
4 s
1 s
8 s

$A c c o r d i n g t o q u e s t i o n . . . . . . . . . . . . . . . . . . . \int_{6.25}^{0} (\frac{d v}{\sqrt{v}}) = - 2.5 \int_{0}^{t} d t \Rightarrow {∣ ∣ 2 \sqrt{v} ∣ ∣}_{6.25}^{0} = - 2.5 t \Rightarrow 2. \sqrt{6.25} = 2.5 t ∴ t = 2 s s o t h e c o r r e c t o p t u o n i s A$