Question

Find velocity of particle when its acceleration is equal to zero, where acceleration (a)=t^2-4t-32
(Given that initially particle is at rest and t is time).

A)640/3 B)213 C)-640/3 D)None of these​

Answer 1

Explanation:

V=(-640/3 )m/s okkkkkkkkkkkk

Answer 2

Answer:

Option C $\frac{-640}{3}$   is correct.

Explanation:

• time when acceleration is zero,

$a = 0\\ t^{2} -4t -32 = 0\\t^{2} -8t + 4t -32 =0\\t(t-8)+4(t-8)=0\\(t+4)(t-8)=0\\t=-4,t=8$

Time can never be negative ,so value ot t = -4 is neglected, only t = 8 is considered.

• $a =\frac{dv}{dt}$

       $dv = adt$

$dv= (t^{2} -4t -32)dt$

upon integration, at

$t= 0,v=0$ and $t = 8 ,v =v$

$\int\limits^v_0 {} \, dv = \int\limits^0_8 {} \,( t^{2} -4t -32)dt\\v-0 = \frac{8^{3} }{3} -2 .8^{2} -32.8\left \ {$

upon simplification,

$v= \frac{-640}{3} \frac{m}{s^{2} }$

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