A sprinter starts from rest and accelerates uniformly for 3 seconds until she reaches a velocity of 10 m s⁻¹ - Leaving Cert Physics - Question b - 2016

Since the sprinter starts from rest and reaches a velocity of 10 m s⁻¹ in 3 seconds, we can determine the acceleration using the formula: $a = \frac{\Delta v}{\Delta t} = \frac{10 \, \text{m s}^{-1} - 0}{3 \, \text{s}} = \frac{10}{3} \, \text{m s}^{-2} \approx 3.33 \, \text{m s}^{-2}$ The graph will show a straight line with this slope for the first 3 seconds.

After the initial acceleration phase, the sprinter runs at a constant velocity of 10 m s⁻¹ for 6 seconds. On the graph, this will appear as a horizontal line at the level of 10 m s⁻¹, extending from 3 seconds to 9 seconds.

The question states that she decelerates after 6 seconds of running at constant velocity. If we assume she comes to rest, then the deceleration can be illustrated as a descending line on the graph. This line can be drawn from the point at (9 s, 10 m s⁻¹) back down to (final time, 0 m s⁻¹).