Question

If the motion of a particle is governed by equation (s=2t^3-3t^2+2t+2). Find the position, velocity and acceleration of the particle at time=2 sec.


Pls urgently pls answer!!!

Answer 1

Hii its tom85

s=2t^3-3t^2+2t+2)

at t= 2

s= 16 - 12 + 4 + 2 = 10m

at t=2

v = ds/dt= 6t^2-6t+2= 24 - 12 +2 = 14m/s

at t=3s

a= dv/dt= 12t - 6 = 6m/s^2

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hope it may helps you

Answer 2

The position, velocity and acceleration of the particle at = 2 sec is 10 m, 14 m/s and 18 m/s².

Explanation:

The motion of a particle is governed by equation as follows :

$s=2t^3-3t^2+2t+2$

Position at t = 2 s

$s=2(2)^3-3(2)^2+2(2)+2\\\\s=10\ m$

Velocity,

$v=\dfrac{ds}{dt}\\\\v=\dfrac{d(2t^3-3t^2+2t+2)}{dt}\\\\v=6t^2-6t+2$

At t = 2 s

$v=6(2)^2-6(2)+2\\\\v=14\ m/s$

Acceleration,

So, the

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6t^2-6t+2)}{dt}\\\\a=12t-6$

At t = 2 s

$a=12(2)-6\\\\a=18\ m/s^2$

So, the position, velocity and acceleration of the particle at = 2 sec is 10 m, 14 m/s and 18 m/s².

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