find the acceleration when the body is instantaneously at rest

bumblebee123

Junior Member
I don't know if I've done the right method, even though I have the right answer. Can anyone check my method?

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!

Subhotosh Khan

Super Moderator
Staff member
I don't know if I've done the right method, even though I have the right answer. Can anyone check my method?

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!
Yes...

Dr.Peterson

Elite Member
One small comment:

There is nothing in the problem as stated that prevents t from being negative. Commonly we take t=0 as the start of motion, but that is not always true; for example, I could write an equation for the position of the moon starting at midnight, but that doesn't mean the moon wasn't there, or wasn't moving, before midnight. We just need to put t=0 somewhere. In your case, you aren't told the motion starts at O, only that it passes through O (at time t=0).

But try taking t=-3!

Jomo

Elite Member
I don't know if I've done the right method, even though I have the right answer. Can anyone check my method?

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!
Helpful suggestion (I hope!): When factoring 3t^2 + 6t - 9 I would think it would be easier for if you factor out the 3 first and get 3(t^2 + 2t - 3). Then continue factoring.