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A train is travelling at 10 m s⁻¹ on a straight horizontal track - Edexcel - A-Level Maths Mechanics - Question 5 - 2005 - Paper 1
Question 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 ... show full transcript
Worked Solution & Example Answer:A train is travelling at 10 m s⁻¹ on a straight horizontal track - Edexcel - A-Level Maths Mechanics - Question 5 - 2005 - Paper 1
Step 1
(a) Sketch a speed-time graph to show the motion of the train.
Answer
To sketch the speed-time graph, we note the following points:
- The train starts at a speed of 10 m s⁻¹ and decelerates to 3 m s⁻¹ over the first 12 seconds.
- The graph will show a linear decrease from 10 m s⁻¹ to 3 m s⁻¹ over this time interval.
- It then travels at a constant speed of 3 m s⁻¹ for another 15 seconds.
- Finally, it decelerates from 3 m s⁻¹ to 0 m s⁻¹ (at rest) until it reaches the signal.
- The graph is characterized by three segments: a declining line, a horizontal line, and another declining line.
Step 2
(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.
Answer
To find the distance travelled during the deceleration from 10 m s⁻¹ to 3 m s⁻¹ over 12 seconds, we can use the formula for distance:
ext{Distance} = rac{1}{2} (u + v) t
Where:
- (initial speed)
- (final speed)
Substituting the values:
ext{Distance} = rac{1}{2} (10 + 3) imes 12 = rac{1}{2} imes 13 imes 12 = 78 ext{ m}
Thus, the distance travelled is 78 m.
Step 3
(c) Find the total time from the moment when the brakes are first applied to the moment when the train comes to rest.
Answer
The total time can be calculated by adding the time spent in each phase:
- From 12 seconds (deceleration from 10 m s⁻¹ to 3 m s⁻¹).
- Then 15 seconds at 3 m s⁻¹ where the train travels with constant speed.
- Finally, calculate the time taken to come to rest after applying brakes:
- The distance remaining to the signal after 27 seconds is:
- Total distance to the signal is 135 m, and the distance covered in the first 27 seconds is 78 m + (15 * 3) = 123 m.
- Remaining distance is: 135 m - 123 m = 12 m.
- Using the equation for slowing down:
- If it decelerates uniformly to rest, using the formula for distance:
- Distance = (1/2) * (initial speed) * (time)
- Let time taken to stop = , then:
- 12 ext{ m} = rac{1}{2} * 3 * t
- Thus,
Therefore, the total time is: