Question

if velocity V as a function of t is equals to for 4t cube + 2 T square + 5t find the displacement travelled by the body in first 2 seconds​

Answer 1

Step-by-step explanation:

$v(t) = 4 {t}^3 + 2 {t}^{2} + 5t$

Differentiate with respect to t we get

$s = \frac{dv(t)}{dt} = \frac{d}{dt}(4 {t}^3 + 2 {t}^{2} + 5t) =$

$12 {t}^{2} + 4t + 5$

initial displacement at t=0 =>5m.

So the total displacement at t=2 seconds =>

$12 {(2)}^{2} + 4(2) + 5 = 48 + 8 + 5 = 61$

Hence the displacement travelled by the body in the first 2 seconds =>61-5=>56m. Hope it helps you.