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So I had my FSMQ exam earlier this year. As a part of the curriculum for this exam we had to learn these rules of SUVAT:

$$v = u + at$$ $$s = ut + \frac{1}{2}at^2$$ $$v^2 = u^2 + 2as$$ $$s = \left( \frac{u + v}{2}\right) t$$ $$a = \frac{dv}{dt}$$ $$v=\frac{ds}{dt}$$ The issue is then when I came to the question, I didn't know how to apply these formula to get the answer, try as I might, there didn't seem to be a way, so I left the question and moved on. However, I would like to know how I could have solved the question as my mum (aka my maths teacher) wasn't really sure and had her hands full teaching the year $10$ s. The question itself is phrased as follows:

Two cars are initially at rest facing in the same direction on a straight road. Car $A$ is $100$m ahead of car $B$. The two cars start from rest at the same moment. Car $A$ moves with a constant acceleration of $1.5$ m s$^{-2}$ and Car $B$ moves with a constant acceleration of $2$ m s$^{-2}$. Find

(I) the distance that car $B$ travels before it overtakes car $A$

(II) the speed of car $B$ at the moment it overtakes car $A$

Maybe there are some other formula I didn't know about, but in my state of understanding I couldn't find the solution despite trying several of the SUVAT equations.

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Gladiator Kittens
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  • In questions like this it's important to have a clear picture of what's going on before you look at equations. Suppose car $B$ overtakes $A$ at time $T$, and that up to this time, it has traveled a distance $s_B$. Since the two cars have met at the time $T$, in the same time, $A$ must have traveled a distance of $s_A = s_B-100$. Next, for each car, you have an expression for how far it travels in time $T$ from the equations above $(2s_B^2 = 2T, 2s_A^2 = 1.5T)$. Now you have three equations and three unknowns, eliminate away. – stochasticboy321 Aug 18 '16 at 09:38
  • @stochasticboy321 how did you get to the last equation in brackets? I can follow the rest of what your saying though :) – Gladiator Kittens Aug 18 '16 at 09:56
  • Those are the equation $s^2 = u^2 + at/2$ evaluated separately for cars $A$ and $B$ (note that $u_A = u_B = 0$, and $a_A = 1.5, a_B = 2$). – stochasticboy321 Aug 18 '16 at 10:08
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    @stochasticboy321 I didn't know that one exists but I used$$ s = ut + \frac{1}{2}at^2$$ and that seems to have worked, don't know if I oversimplified somewhere because the answer numbers are surprisingly round ( I got 100m for the distance and 20m/s for the speed) – Gladiator Kittens Aug 18 '16 at 10:10
  • Oh, wait, this is my error. Those should be $s_A/s_B$ instead of $s_A^2 /s_B^2$. Apologies. Your method in the comment above is correct. – stochasticboy321 Aug 18 '16 at 10:13
  • @stochasticboy321 thanks so much for the help, definitely wouldn't have gotten the answer on my own, had a little bit of a "duh that's how" moment though. Glad to finally have the question off of my chest, especially after getting my exam result this morning of an A, happy the missing 6 marks didn't drag me down :) if you could add an answer so I can close the question that would be appreciated,even if it's just a copy paste of the comment so I can accept it – Gladiator Kittens Aug 18 '16 at 10:16
  • Grand :). Another thing that might be helpful with these problems is to draw a figure. Things like $s_A = s_B -100$ tend to pop right out when you do that. – stochasticboy321 Aug 18 '16 at 10:18
  • @stochasticboy321 I think I got $$sA = sB - 100$$ but I didn't know what to do with it once I had it, exam pressure makes people do silly things – Gladiator Kittens Aug 18 '16 at 10:20

1 Answers1

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\begin{align*} s_A &= u_A t+\frac{1}{2}a_A t^2 \\ &= 0(t)+\frac{1}{2}(1.5)t^2 \\ s_B &= u_B t+\frac{1}{2}a_B t^2 \\ &= 0(t)+\frac{1}{2}(2)t^2 \\ s_B &= s_A + 100 \\ t^2 &= 0.75t^2 + 100 \\ t^2 &= 400 \\ t &= 20 \text{ s} \\ v_B &= u_B+a_B t \\ &= 2(20) \\ &= 40 \text{ m s}^{-1} \end{align*}

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