An object, moving with a speed 6.25ms−1, is decelerated at a rate given by dvdt=−2.5√v where v is intantaneous velocity. The time taken by the object to come to rest would be:
2s
4s
8s
1s
A
4s
B
8s
C
2s
D
1s
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Solution
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dvdt=2.5√v
dv√v=−2.5dt
∫06.25v−1/2dv=−2.5∫t0dt
−2.5[t]t0=[2v1/2]06.25
t=2s
dvdt=−2.5√v
∫06.25V−1/2dv=−2.5∫t0dt
−2.5[t]t0=[2v1/2]06.25
t=2seconds
Hence (A) option is corrcet
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