M1 June 2005 Q5
5. A train is travelling at 10 m s\(^{-1}\) on a straight horizontal track. The driver sees a red signal 135 m ahead and immediately applies the brakes. The train immediately decelerates with constant deceleration for 12 s, reducing its speed to 3 m s\(^{-1}\). The driver then releases the brakes and allows the train to travel at a constant speed of 3 m s\(^{-1}\) for a further 15 s. He then applies the brakes again and the train slows down with constant deceleration, coming to rest as it reaches the signal.
(a) Sketch a speed-time graph to show the motion of the train, (3)
(b) Find the distance travelled by the train from the moment when the brakes are first applied to the moment when its speed first reaches 3 m s\(^{-1}\). (2)
(c) Find the total time from the moment when the brakes are first applied to the moment when the train comes to rest. (5)
| Scheme | Marks |
|---|---|
| Shape \(0 \lt t \lt 12\) | B1 |
| Shape \(t \gt 12\) | B1 |
| Figures | B1 |
| (3) |
| Scheme | Marks |
|---|---|
| Distance in 1st 12 s \(= \tfrac{1}{2} \times (10 + 3) \times 12\) or \((3 \times 12) + \tfrac{1}{2} \times 3 \times 7\) | M1 |
| \(= 78\) m | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| either distance from \(t = 12\) to \(t = 27 = 15 \times 3 = 45\) \(\therefore\) distance in last section \(= 135 - 45 = 12\) m | B1ft |
| \(\tfrac{1}{2} \times 3 \times t = 12\), | M1 A1ft |
| \(\Rightarrow t = 8\) s | A1 |
| hence total time \(= 27 + 8 = 35\) s | A1 |
| (5) | |
| (10 marks) |
Notes
or
| Scheme | Marks |
|---|---|
| Distance remaining after 12 s \(= 135 - 78 = 57\) m | B1ft |
| \(\tfrac{1}{2} \times (15 + 15 + t) \times 3 = 57\) | M1 A1ft |
| \(\Rightarrow t = 8\) | A1 |
| Hence total time \(= 27 + 8 = 35\) s | A1 |