Two trains A and B run on parallel straight tracks. Initially both are at rest in a station and level with each other. At time $t = 0$, A starts to move. It moves with constant acceleration for 12 s up to a speed of 30 m s$^{-1}$, and then moves at a constant speed of 30 m s$^{-1}$. Train B starts to move in the same direction as A, when $t = 40$, where $t$ is measured in seconds. It accelerates with the same initial acceleration as A, up to a speed of 60 m s$^{-1}$. It then moves at a constant speed of 60 m s$^{-1}$. Train B overtakes A when $t = T$. (a) Sketch, on the same diagram, the speed-time graphs of both trains for $0 \leq t \leq T$. (b) Find the value of $T$.

A-Level Edexcel Maths Mechanics Constant Acceleration - 1D: Two trains A and B run on parall

Two trains A and B run on parallel straight tracks - Edexcel - A-Level Maths Mechanics - Question 7 - 2003 - Paper 1

Question 7

Two trains A and B run on parallel straight tracks. Initially both are at rest in a station and level with each other. At time $t = 0$, A starts to move. It moves wi... show full transcript

Worked Solution & Example Answer:Two trains A and B run on parallel straight tracks - Edexcel - A-Level Maths Mechanics - Question 7 - 2003 - Paper 1

Step 1

Sketch, on the same diagram, the speed-time graphs of both trains for $0 \leq t \leq T$

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Answer

To sketch the speed-time graphs for both trains:

Train A:
- Starts from rest and accelerates from $t = 0$ to $t = 12$ seconds.
- Final speed at $t = 12$ s is 30 m s $^{-1}$ , which is maintained thereafter.
- Graph linearly rises from (0,0) to (12,30) and remains horizontal from (12,30).
Train B:
- Starts moving at $t = 40$ seconds with the same initial acceleration as Train A.
- It accelerates for 24 seconds (12 seconds + 12 seconds extra) to reach a speed of 60 m s $^{-1}$ .
- The speed-time graph will remain horizontal at 60 m s $^{-1}$ after reaching maximum speed.

Overall, the sketch will show two lines:

Train A's line from (0,0) to (12,30) then flat to $(T,30)$ .
Train B starts from (40,0), accelerates up to $(64,60)$ then remains flat to $(T,60)$ .

Step 2

Find the value of $T$

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Answer

To find the value of $T$ , we can calculate the distance traveled by both trains and set them equal when Train B overtakes Train A:

Distance moved by Train A:
- For the first 12 seconds: $d_A = \frac{1}{2} \times 12 \times 30 = 180 \text{ m}$
- Then it maintains a constant speed of 30 m s $^{-1}$ for $(T - 12)$ seconds: $d_A = 180 + 30 \times (T - 12) \text{ m}$
Distance moved by Train B:
- Train B accelerates for 24 seconds at the same rate as Train A: $d_B = \frac{1}{2} \times 24 \times 60 = 720 \text{ m}$
- After this, it travels at 60 m s $^{-1}$ for $(T - 64)$ seconds: $d_B = 720 + 60 \times (T - 64) \text{ m}$

Now set the distances equal: $180 + 30 \times (T - 12) = 720 + 60 \times (T - 64)$

Solving for $T$ : $180 + 30T - 360 = 720 + 60T - 3840$ $30T - 60T = 720 - 180 - 3840 + 360$ $-30T = -3240$ $T = 108 ext{ seconds}$

The value of $T$ is 98 seconds as given.

Two trains A and B run on parallel straight tracks - Edexcel - A-Level Maths Mechanics - Question 7 - 2003 - Paper 1

Worked Solution & Example Answer:Two trains A and B run on parallel straight tracks - Edexcel - A-Level Maths Mechanics - Question 7 - 2003 - Paper 1

Sketch, on the same diagram, the speed-time graphs of both trains for $0 \leq t \leq T$

Find the value of $T$

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