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AQA GCSE · Question 07.3 · Forces
Determine the maximum deceleration of the train.
Determine the maximum deceleration of the train.
How to approach this question
1. Deceleration is the gradient of the velocity-time graph. Maximum deceleration occurs where the graph is steepest.
2. Identify the section of the graph with the steepest negative slope.
3. Calculate the gradient for this section using: gradient = (change in y) / (change in x) = (change in velocity) / (time taken).
4. Read the velocity and time values from the start and end of this section accurately.
Full Answer
1. Maximum deceleration occurs where the graph is steepest, between t=650 s and t=800 s.
2. Deceleration = gradient = change in velocity / time taken.
3. Change in velocity = 58 m/s - 20 m/s = 38 m/s.
4. Time taken = 800 s - 650 s = 150 s.
5. Deceleration = 38 / 150 = 0.253 m/s² (to 3 s.f.)
Deceleration is the rate of change of velocity, which is the gradient of the velocity-time graph. The maximum deceleration corresponds to the steepest downward-sloping section of the graph.
Looking at the graph, the steepest section is between t = 650 s and t = 800 s.
- At t₁ = 650 s, the velocity v₁ = 58 m/s.
- At t₂ = 800 s, the velocity v₂ = 20 m/s.
Gradient = (change in velocity) / (change in time) = (v₂ - v₁) / (t₂ - t₁)
Gradient = (20 - 58) / (800 - 650)
Gradient = -38 / 150
Gradient = -0.2533... m/s²
Deceleration is the positive value of this gradient.
Maximum deceleration = 0.253 m/s² (to 3 significant figures).
Common mistakes
✗ Calculating the area instead of the gradient.\n✗ Misreading the values from the axes.\n✗ Choosing the wrong section of the graph (the less steep part).\n✗ Giving the answer as a negative number (acceleration is negative, deceleration is positive).
Practice the full AQA GCSE Physics Higher Tier Paper 2
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