# A Level Maths Question- ​A particle P moves along a straight line such that at time t

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#1
I really struggle with mechanics any advice on how to approach this question would be appreciated.

A particle P moves along a straight line such that at time t seconds, t ≥ 0, after leaving the point O on the line, the velocity, v m s–1, of P is modelled as

v= (7-2t)(t+2)

(a) Find the value of t at the instant when P stops accelerating.

(b) Find the distance of P from O at the instant when P changes its direction of motion.
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1 year ago
#2
Here’s some hints

a) As acceleration stops acceleration must be variable so you can’t use SUVAT and must differentiate instead to find acceleration. So the first step is to expand.

b) To find s you need to integrate

Let me know if you need more help 0
#3
For (a), I expanded it to get -2t^2 + 3t + 14. And I then differentiated to get -4t+3. Do I just do -4t + 3 = 0 to calculate t or am I oversimplifying it? I really appreciate the help, thanks 2
1 year ago
#4
(Original post by 4Lozza)
For (a), I expanded it to get -2t^2 + 3t + 14. And I then differentiated to get -4t+3. Do I just do -4t + 3 = 0 to calculate t or am I oversimplifying it? I really appreciate the help, thanks Yep! The derivative of velocity is acceleration, so the time at which P stops accelerating is the same as dv/dt = 0.
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#5
(Original post by unplayable)
Yep! The derivative of velocity is acceleration, so the time at which P stops accelerating is the same as dv/dt = 0.
So t just equals 3/4?

Then what do I do for part (b)? I integrated it to get -2/3t^3 + 3/2t^2 +14t + c but I don't know what to do after that. Thanks for the help Last edited by 4Lozza; 1 year ago
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1 year ago
#6
(Original post by 4Lozza)
So t just equals 3/4?

Then what do I do for part (b)? Thanks for the help
Yeah don't forget to add units.

Use your expanded brackets and integrate. To find the value for c find t when v=0 (i think)

edit: sorry to find c use the equation for s at t=0 --> s=0 also at t=0
Last edited by Hellllpppp; 1 year ago
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1 year ago
#7
(Original post by Hellllpppp)
Yeah don't forget to add units.

Use your expanded brackets and integrate. To find the value for c find t when v=0 (i think)

edit: sorry to find c use the equation for s at t=0 --> s=0 also at t=0
I would do it ever-so-slightly differently, but they're essentially the same thing and I think we get the same answer. First, work out the time at which P changes direction (i.e. v = 0), then do a definite integral of v(t) with a lower bound of 0 and an upper bound of t when v = 0.
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#8
(Original post by unplayable)
I would do it ever-so-slightly differently, but they're essentially the same thing and I think we get the same answer. First, work out the time at which P changes direction (i.e. v = 0), then do a definite integral of v(t) with a lower bound of 0 and an upper bound of t when v = 0.
So I worked out that t = 7/2 when v = 0 (please correct me if I’m wrong). And I used integration to get -2/3t^3 + 3/2t^2 +14t + c. What am I meant to do now? And how do I find C? Thanks for all the help 😃
Last edited by 4Lozza; 1 year ago
0
1 year ago
#9
(Original post by 4Lozza)
So I worked out that t = 7/2 when v = 0 (please correct me if I’m wrong). And I used integration to get -2/3t^3 + 3/2t^2 +14t + c. What am I meant to do now? And how do I find C? Thanks for all the help 😃
c would be 0 as distance from 0 when t = 0 is 0.

Now sub in t = 3.5 into your integrated equation.
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#10
c would be 0 as distance from 0 when t = 0 is 0.

Now sub in t = 3.5 into your integrated equation.
I substituted it in and I got 38.79. Is that correct? and what do I do next? thanks so much for all the help
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