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M3 Help! Stuck on a question.

In the Heinemann Modular Mathematics M3 book, question 13 of Exercise 1A goes as follows:

A particle PP moves in a straight line with acceleration, at time tt seconds, proportional to (t+t0)−3(t+t_0)^{-3}, where t0t_0 is positive and constant. The initial speed of PP at t=0 t = 0 is u ms−1u\ ms^{-1}. Show that the speed of PP approaches at limiting value as t→∞t \to \infty. Given that this limiting value is 2u ms−12u\ ms^{-1}. show that at time tt seconds the particle will have travelled a distance ut(2t+t0)(t+t0)\frac{ut(2t+t_0)}{(t+t_0)} meters.

So far I've gone so far as to make these arguments.

When t=t,a∝(t+t0)−3\text{When } t=t, a \propto (t+t_0)^{-3}
⇒a=k(t+t0)−3\Rightarrow a = k(t+t_0)^{-3}
When t=0,v=u\text{When } t=0, v=u

v= ∫a dt= k∫(t+t0−3) dt=−k(t+t0)−22+c=−k2(t+t0)2+cv =\displaystyle\ \int \mathrm{a}\,\mathrm{d}t =\displaystyle\ k\int (\mathrm{t+t_0}^{-3})\,\mathrm{d}t = -k\frac{(t+t_0)^{-2}}{2} +c =-\frac{k}{2(t+t_0)^{2}} +c

 lim⁡t→∞u=−k2(t0)2\displaystyle\ \lim_{t \to \infty} u = -\frac{k}{2(t_0)^2}

I'm kind of stuck after that.
Original post by kelechi96
...


When you get

v=−k2(t+t0)2+cv = -\dfrac{k}{2(t+t_0)^{2}} +c

You should use the initial conditions to find the constant of integration.

I'm not exactly sure if this is Edexcel, but if it is, 1A Q13 doesn't exist

Original post by kelechi96

 lim⁡t→∞u=−k2(t0)2\displaystyle\ \lim_{t \to \infty} u = -\frac{k}{2(t_0)^2}

I'm kind of stuck after that.


Don't know where that last bit has come from.

You need to use the fact that when t=0, velocity = u, and as t goes to infinity, velocity goes to 2u, to determins k and c.

This will then give you a formula for v.

Then integrate with limits t and 0 to find the distance travelled.
Original post by ghostwalker
Don't know where that last bit has come from.

You need to use the fact that when t=0, velocity = u, and as t goes to infinity, velocity goes to 2u, to determins k and c.

This will then give you a formula for v.

Then integrate with limits t and 0 to find the distance travelled.


Have you completed the question?

I got x=ut(2t−t0)t+t0 x = \frac{ut(2t-t_0)}{t+t_0} \ but this disagrees with the OPs answer which is x=ut(2t+t0)t+t0x = \frac{ut(2t+t_0)}{t+t_0}

:redface:
(edited 10 years ago)
Original post by Khallil
Have you completed the question?

I got x=ut(2t−t0)t+t0 x = \frac{ut(2t-t_0)}{t+t_0} \ but this disagrees with the OPs answer which is x=ut(2t+t0)t+t0x = \frac{ut(2t+t_0)}{t+t_0}

:redface:


Yes, I agree with the OP's answer.

I got c=2uc=2u and k=2ut02k=2ut_0^2
(edited 10 years ago)

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