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# A particle is moving with accleration a= 5t+4 ..Then its velocity after 4 second is a. 44m/s b. 48 m/s c. 56 m/s d. 60m/s

Sir ji help in this question and in video lecture u will directly  calculate the accleretion value but sir there are many other students like me who will have only subjects like P.C.B. So they are not undestand so kindly tell us how we are able to solve this types of questions in detail and and clear my above doubt also.

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$a = \frac{\mathrm{d} v}{\mathrm{d} t}$

$\mathrm{d} v = (5t+4)\mathrm{d} t$

$\int_{u}^{v}\mathrm{d} v = \int_{0}^{4}(5t+4)\mathrm{d} t$

$\Rightarrow v-u =[\frac{5}{2}t^{2}+4t]_{0}^{4}$

$\Rightarrow v-u =5\times \frac{4^{2}}{2}+4\times 4$

$\Rightarrow v-u =56$

Assuming initial velocity is zero, so u = 0.

hence  $v =56$

option (c) is correct

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