The graph shows the speed of a Luas tram as it travels from Stop L to Stop Q - Junior Cycle Science - Question c - 2015

To calculate the distance travelled, we can use the formula:

$ext{Distance} = ext{Speed} imes ext{Time}$
From the graph, the speed between positions M and N is 10 m/s and the time interval is from 40 seconds to 100 seconds, therefore the time taken is 100 - 40 = 60 seconds.
Substituting the values, we have:

$ext{Distance} = 10 ext{ m/s} imes 60 ext{ s} = 600 ext{ m}$

Acceleration can be calculated using the formula:

$a = \frac{v_f - v_i}{t}$
where $v_f$ is the final velocity and $v_i$ is the initial velocity.
From the graph, $v_f = 15 ext{ m/s}$ and $v_i = 10 ext{ m/s}$ between positions N and O. The time taken from N to O is 40 seconds (from 100 to 140 seconds).
Thus, substituting the values, we get:

$a = \frac{15 - 10}{40} = \frac{5}{40} = 0.125 ext{ m/s}^2$

The units of acceleration are meters per second squared (m/s²).

Between positions P and Q, the tram is decelerating, which means it is undergoing negative acceleration as it slows down. The speed decreases from 15 m/s to 0 m/s, indicating that the tram is coming to a stop.