Question

A particle is moving along x axis whose velocity is given by

V = 2t² + 4

Find its acceleration when t = 4 second ????​

Answer 1

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$16 \ m \ sec^{-2}$

Explanation:

$Given \ velocity \ of \ particle \ v=2t^2+4\\\\\\we \ have \ to \ find \ acceleration \ at \ t = 4 \ second\\\\\\we \ know \ that \ acceleration=\frac{change \ in \ velocity}{time}\\\\\\ a=\frac{dv}{dt}\\\\\\a=\frac{d(2t^2+4)}{dt}\\\\\\using \ power \ rule \ of \ differentiation\\\\\\x^n=n.x^{n-1}\\\\\\a=2\times2t^{2-1}+0\\\\\\a=4t\\\\\\putting \ t=4\\\\\\a=4\times4\\\\\\a=16 \ m \ sec^{-2}\\\\\\Thus, \ we \ get \ our \ acceleration \ 16 \ m \ sec^{-2}$

Answer 2

Hola mate

Answer:

We know

a = 4t

substitute value of t

we get

a = 4 × 4 = 16 m/s^2