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Two cars start from rest, from the start line, and accelerate uniformly along a racetrack running perpendicular to the start line. After 5 seconds the first car is $30\,\mathrm m$ in front of the second car. How far is it in front after another $5$ seconds?

Using a velocity time graph, I have been able to calculate the differences between the velocities. $v_1$ denotes the velocity of the faster car and $v_2$ that of the slower one. $0.5 \times 5 \times v_1 = 2.5 v_1$,

$0.5 \times 5 \times v_2 = 2.5 v_2$,

$2.5(v_1-v_2) = 30$,

$v_1 - v_2 = 12$

Using the equation $(v-u)/t$, I used a similar method to find that $a_1 - a_2 = 2.4$ ($a$ denotes acceleration). However, I am unsure as to how to use this information to find the solution to the question. If you can help me, I would be grateful.

Izaak van Dongen
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AOD
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1 Answers1

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In this case I think a nice quick way is to think about it as follows: since we're interested in the displacement of the faster car relative to the slower car (and that's all we're given information about) we might as well measure everything relative to the slower car.

The faster car still has uniform acceleration relative to the slower car ($a_1 - a_2$ if you like), which we will just call $a$. Both initial velocities were $0$, so the initial velocity of the faster car relative to the slower car is still $0$. Now we know $s$ when $t = 5$, and $u$. We would like to find out $s$ when $t = 10$.

This is given by the equation $s(t) = ut + \frac 12 at^2$, which simplifies a bit because $u = 0$: we just get $s(t) = \frac 12 at^2$. Now note that $s(10) = \frac 12 a(10)^2 = 4 \cdot \frac 12 a(5)^2 = 4 \cdot s(5) = 4 \cdot 30 = 120$. So after $10$ seconds, the second car is $120\,\mathrm m$ ahead.

Izaak van Dongen
  • 3,791