A train with mass $1.8 \times 10^5\,\text{kg}$ is initially stationary in a station. When $t = 0$, it starts to speed up along a straight, horizontal track and attains a speed of $20\,\text{m s}^{-1}$ by $t = 15\,\text{s}$. It then travels at $20\,\text{m s}^{-1}$ for another $10\,\text{s}$. At $t = 25\,\text{s}$, the driver puts on the brakes and the resistive force on the train makes it slow down uniformly to rest over a further $24\,\text{s}$. Fig. 4.1 is an unfinished distance-time graph for this trip.
Fig. 4.1 plots distance/m against $t/\text{s}$.
(a(i))[1]
Finish Fig. 4.1 by adding a line that shows the train’s motion from $t = 15\,\text{s}$ to $t = 25\,\text{s}$.
(a(ii))[1]
Finish Fig. 4.1 by drawing a curve that shows the train’s motion from $t = 0$ to $t = 15\,\text{s}$.
(b)[3]
Calculate the kinetic energy of the train over the interval $t = 15\,\text{s}$ to $t = 25\,\text{s}$.
(c(i))[2]
Define work done
(c(ii))[1]
From Fig. 4.1, determine how far the train travels during deceleration.
(c(iii))[2]
Calculate the resultant force acting on the train as it decelerates.
Worked solution & mark scheme
This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “A straight-line segment from (15 s, 120 m) to the end of the provided line” …