Question

A tram leaves a station and accelerates for $2$ minutes until it reaches a speed of $12$ metres per second.
It continues at this speed for $1$ minute
It then declerates for $3$ minutes until it stops at teh next station.
The diagram shows the speed-time graph for this journey.
calculate the distance, in metres, between the two stations.

Answer 1

1) Consider first part of velocity-time curve.

As velocity-time curve is linear, acceleration is constant and magnitude of acceleration is slope of velocity-time curve

$\therefore a=\frac{12}{2}$

$\therefore a=6\frac{m}{s^2}$

Using equation of motion with constant acceleration,

$s_1=ut+\frac{1}{2}a{t}^2$

$s_1$ is distance covered in first 2 seconds

$u=0$ is initial velocity for first part

$t=2$ is time for first part

$a=6$ is constant acceleration for first part

$\therefore s_1=0+\frac{1}{2}\left(6\right){\left(2\right)}^2$

$\therefore s_1=3\times 4$

$\therefore s_1=12m$

2) Consider second part of curve where velocity is constant.

Distance covered by tram from $t=2$ to $t=3$ is given by,

$s_2=v\times t$

$\therefore s_2=12\times 1$

$\therefore s_2=12m$

3) Consider third part of the curve.

As velocity is decreasing, acceleration will be negative and equal to slope of velocity time curve.

$\therefore a=-\frac{12}{3}$

$\therefore a=-4\frac{m}{s^2}$

Using equation of motion with constant acceleration,

$s_3=ut+\frac{1}{2}a{t}^2$

$\therefore s_3=(12\times 3)+\frac{1}{2}(-4){\left(3\right)}^2$

$\therefore s_3=36-2\times 9$

$\therefore s_3=36-18$

$\therefore s_3=18m$

Thus, total distance covered by the tram in 6 seconds is given by,

$s=s_1+s_2+s_3$

$\therefore s=12+12+18$

$\therefore s=42m$