#### bumblebee123

##### Junior Member

- Joined
- Jan 3, 2018

- Messages
- 196

**question:**a body is moving in a straight line which passes through a fixed point

*O.*The displacement, x metres, of the body from O at time t seconds is given by

x= t^3 + 3t^2 - 9t

**a)**find an expression for the velocity, v m/s, at the time t seconds

I managed to do this: v = 3t^2 + 6t - 9

**b)**find the acceleration when the body is instantaneously at rest

I thought this would be when the velocity = 0

so 0 = 3t^2 + 6t - 9

0 = ( 3t - 3 ) ( t + 3 )

t = 1 or t= -3

however, t must = 1 because it can't be negative

so if I put this into the acceleration equation: a = 6t + 6

a = 6 + 6

a = 12 m/s^-2 which is correct

did I do it right? thanks!