Question

y =( -2/3)t² +2t+ 5 is position. find time when body comes to rest

Answer 1

Answer:

1.5 sec

Explanation:

Given:

$\mathsf{y =( -2/3){t}^{2} +2t+ 5}$

is the position of the body

To Find

time after when it comes to rest

Solution

$\mathsf{y =( -2/3){t}^{2} +2t+ 5}$

Differentiating on both sides, w.r.t Time

$\mathsf{\dfrac{dy}{dt} =\dfrac{d[( \dfrac{-2}{3}){t}^{2}+2t+ 5]}{dt}}$

$\mathsf{v = \dfrac{-4}{3}t+2}$

V= 0

$\mathsf{0 = \dfrac{-4}{3}t+2}$

$\mathsf{-2 = \dfrac{-4}{3}t}$

$\mathsf{-2 \times \dfrac{-3}{4} = t}$

$\mathsf{ \dfrac{3}{2} = t}$

$\mathsf{t = 1.5 sec}$