M1 January 2012 Q6
6. A car moves along a straight horizontal road from a point \(A\) to a point \(B\), where \(AB = 885\) m. The car accelerates from rest at \(A\) to a speed of 15 m s\(^{-1}\) at a constant rate \(a\) m s\(^{-2}\). The time for which the car accelerates is \(\tfrac{1}{3}T\) seconds. The car maintains the speed of 15 m s\(^{-1}\) for \(T\) seconds. The car then decelerates at a constant rate of 2.5 m s\(^{-2}\) stopping at \(B\).
(a) Find the time for which the car decelerates. (2)
(b) Sketch a speed-time graph for the motion of the car. (2)
(c) Find the value of \(T\). (4)
(d) Find the value of \(a\). (2)
(e) Sketch an acceleration-time graph for the motion of the car. (3)
| Scheme | Marks |
|---|---|
| \(v = u + at \ \Rightarrow\ 0 = 15 - 2.5t\) | M1 |
| \(t = 6\ \ (\text{s})\) | A1 |
| (2) |

| Scheme | Marks |
|---|---|
| Shape | B1 |
| 15, \(T\) | B1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}15\left(\dfrac{4}{3}T + 6 + T\right) = 885\) ft their 6 | M1 A1ft |
| \(\dfrac{7}{3}T = 118 - 6\) | |
| \(T = 112 \times \dfrac{3}{7} = 48\) | M1 A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(a = \dfrac{15}{\tfrac{1}{3}T} = \dfrac{15}{16}\), 0.9375, 0.938, 0.94 | M1 A1 |
| (2) |

| Scheme | Marks |
|---|---|
| 3 horizontal lines | B1 |
| Correctly placed; no cts vert line | B1 |
| \(-2.5\), ft their \(\tfrac{15}{16}\) | B1 |
| (3) | |
| (13 marks) |