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.
Best Answer
\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*}