M1 Question Please help.

The question: A particle P is moving along the x-axis with constant deceleration 3ms-2. At time t=0s, P is passing through the origin O and is moving with speed ums-1 in the direction of x increasing. At time t=8s, P is instantaneously at rest at the pout A. Find:

A) The value of u
B) The distance OA
C) The times at which P is 24m from A.

I worked out A (got 24ms-1) and B (got 96m) which are correct according to the book. I just can't work out C. I think I know what I have to do, but I can't seem to get the right answers.

This is what I have done:

Using the equation s=ut+1/2at^2 (it says times, so I assume you make a quadratic equation from this.

s=24m
a=-3ms-2

Therefore I have 24=ut -3/2t^2

I wasn't too sure how to get u, so I used the fact that s=24m, and a=-3ms-2. Then I used v=0ms-1. I then used v^2=u^2+2as

v^2 - 2as = u^2

This gives u=12ms-1.

So then I have 24=12t -3/2t^2

Have I done this at all correct so far?

Thanks

Oh so I still use 24ms-1 then. But then surely s=96m, because u (in part A) was starting at O, an in part C I am starting from 72m away from O (24m from A)?

Sorry if that doesn't make sense

I think you would use original value for u and then use s=72m

Oh so I still use 24ms-1 then. But then surely s=96m, because u (in part A) was starting at O, an in part C I am starting from 72m away from O (24m from A)?

Sorry if that doesn't make sense

I see. I did consider doing that but then just ended up totally confused.

If I do that I would get

72=24t - 3/2t^2

Re-arranged: 3/2t^2 - 24t + 72 = 0

Is that correct?

hmmmmm
You have 2 choices u=24 and then s = OA-24


Or you can start again from A in which case u=0 and s = 24 not forgetting to add the 8 sec from the start

hmmmmm
You have 2 choices u=24 and then s = OA-24


Or you can start again from A in which case u=0 and s = 24 not forgetting to add the 8 sec from the start

It would give two real values for t, it is correct in the substitution you've done try solving it and see if the answers are right

Oh so I still use 24ms-1 then. But then surely s=96m, because u (in part A) was starting at O, an in part C I am starting from 72m away from O (24m from A)?

Sorry if that doesn't make sense

I think that you've slightly misinterpreted the question.
Using the information provided in the question, and also the information you have calculated in parts a) and b), I will describe the motion of the particle:

At time $t = 0s{-1}$, $u = 24ms^{-1}$.
It then moves forward with a constant deceleration, so that $a = -3ms^{-2}$.
When the particle has a displacement $s = 96m$, it comes to instantaneous rest, but it is still accelerating in the negative direction.

Twice on this journey, $s = 24m$, the first when it has a positive velocity, the second when it has a negative velocity. Your task is to find out the times at which this occurs.

Here's a hint:


I'm sorry if I've repeated anything you've already said or know.

This way would also work

hmmmmm
You have 2 choices u=24 and then s = OA-24


Or you can start again from A in which case u=0 and s = 24 not forgetting to add the 8 sec from the start

Y

Thank you both as well I got it right this time .

Yep I worked it out a couple of minutes ago and I got that, and the solutions work.

Thanks for explanation








Thank you both as well I got it right this time .