Maths: Kinematics

Please help with this question: A particle is projected at a speed of 12ms-1 up a straight inclined track. Whilst on the track the particle experiences a constant acceleration down the track of 5ms-2. Find the length of time that the particle is more than 8 metres from the point of projection.

Do you have an answer for this?
(I have one but not sure it’s right)

No I do not

not sure if this helps as I'm not sure how to do the whole question but to start use the SUVAT equations ->
s = 8
u = 12
v = v
a = 5
t = ?

Ah, shoot. I like to know I’m right before I help..

What have you done so far?

I just wrote the suvat for it but Im still not sure whether gravity has anything to do with it so is a: 5 or 9.81

I just wrote the suvat for it but Im still not sure whether gravity has anything to do with it so is a: 5 or 9.81

It is on a track and the question tells you the constant acceleration is 5m/s DOWN the track.

So did you get 0.44 (3sf) as your final answer

I did not...but that doesn’t mean you’re not correct. Show your working

I did s = ut+1/2at^2
Subbed in the values to get
8 = 12t+2.5t^2
Then did 2.5t^2+12t-8 = 0
Did the quadratic formula and got 0.5933...
So not 0.44 my bad

You’re looking for the length of time it is More than 8 metres away.
Also your acceleration is down the track so should be negative.

(My confusion lies in whether it returns back to the start once it gets to the top point of the track)

It should do. The acceleration will remain.

You’re looking for the length of time it is More than 8 metres away.
Also your acceleration is down the track so should be negative.

(My confusion lies in whether it returns back to the start once it gets to the top point of the track)

I don't think it does, but how did you work it out?

I started by working out the time until it stops moving upwards.

Please tell me more

Have you done that?