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Grade 11,

The velocity time relation of a body starting from rest is given by v = k^2t^3/2 ,where k = (2)^1/2 m^1/2/s^5/4. the distance traversed in 2 seconds is?

The velocity time relation of a body starting from rest is given by v = k^2t^3/2 ,where k = (2)^1/2 m^1/2/s^5/4. the distance traversed in 2 seconds is?

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1 Answers

Shubham Vernekar
15 Points
5 years ago

Given, v=k^2t^(3/2)

where k=(2^0.5)(m^0.5)/(s^5/4)

then, k^2 = (2m)/(s^5/2)

:. v = (2m)(t^3/2)/(s^5/2)

=>ds/dt = (2m)(t^3/2)/(s^5/2)

=>s^5/2 ds = (2m) t^3/2 dt

integrating, from t=0 to t=4 ,

(2/7)s^7/2 = (2m) [(2/5) 4^(5/2)]

=> (7/2) s^7/2 = 25.6m

=> s^(7/2) = (89.6m)

=> s = (89.6m)^(7/2)

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