The graph shows the distance, \(s\) metres, of a car from the car park barrier, \(t\) seconds after moving off, modelled by \(s=6t^{2}-t^{3}\) for \(0 \le t...
Question 1 Report
The graph shows the distance, \(s\) metres, of a car from the car park barrier, \(t\) seconds after moving off, modelled by \(s=6t^{2}-t^{3}\) for \(0 \le t \le 6\), as shown.
- Using the graph, write down the value of \(t\), other than \(0\), at which the car is level with the barrier again. (1)
- Find \(\dfrac{ds}{dt}\). (1)
- Work out the car's velocity at \(t=2\) seconds. (2)
- Find the time at which the car is momentarily at rest, and the distance it has travelled by then. (2)
Answer Details
(a) Reading from the graph, the curve returns to \(s=0\) at \(t=6\) seconds (confirmed algebraically, since \(s=6t^{2}-t^{3}=t^{2}(6-t)\) is zero at \(t=0\) and \(t=6\)). [1 mark]
(b) Differentiating \(s=6t^{2}-t^{3}\) term by term gives \(\dfrac{ds}{dt}=12t-3t^{2}\). [1 mark]
(c) The velocity is the value of \(\dfrac{ds}{dt}\) at the given time. Substituting \(t=2\): \(12(2)-3(2)^{2}=24-12=12\) m/s. [2 marks]
(d) The car is momentarily at rest when its velocity is zero: \(12t-3t^{2}=0\), which factorises as \(3t(4-t)=0\), giving \(t=0\) or \(t=4\). Other than the start, this occurs at \(t=4\) s. The distance travelled by then is \(s=6(4)^{2}-(4)^{3}=96-64=32\) m. [2 marks]
The derivative gives the car's velocity at any instant; where that derivative is zero, the distance-time graph has a turning point, here a local maximum distance from the barrier before the car starts moving back towards it.
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