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Two cars P and Q are moving in the same direction along the same straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 5 - 2010 - Paper 1

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Two cars P and Q are moving in the same direction along the same straight horizontal road. Car P is moving with constant speed 25 m s-1. At time t = 0, P overtakes Q... show full transcript

Worked Solution & Example Answer:Two cars P and Q are moving in the same direction along the same straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 5 - 2010 - Paper 1

Step 1

Sketch, on the same axes, the speed-time graphs of the two cars for the period from t = 0 to the time when they both come to rest at the point X.

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Answer

To create the speed-time graphs for cars P and Q, we must plot their speeds against time.

  1. For Car P:

    • From t = 0 to t = T, Car P moves at a constant speed of 25 m/s.
    • At t = T, Car P begins to decelerate uniformly to rest at X.
    • From t = T to t = 25 s, it will gradually decrease to 0 m/s.
  2. For Car Q:

    • For the first 25 seconds, Car Q moves at a constant speed of 20 m/s.
    • After 25 seconds, it begins to decelerate uniformly and also comes to rest at point X at the same time as Car P.

The graph will show the following features:

  • Both cars start at different speeds.
  • The lines for both cars must intersect on the time axis at the point where they both stop at X.
  • You will have horizontal lines for constant speeds and sloping lines for deceleration, meeting at the point where both cars stop.

Step 2

Find the value of T.

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Answer

To determine T, we use the distances covered by both cars to set up the equations:

  1. For Car Q:

    It travels at 20 m/s for T seconds then decelerates.

    Distance covered by Q before deceleration: dQ=20T+202(25T)=20T+10(25T)=20T+25010T=10T+250d_Q = 20T + \frac{20}{2}(25 - T) = 20T + 10(25 - T) = 20T + 250 - 10T = 10T + 250

    Setting this equal to the distance to X: 10T+250=80010T + 250 = 800 10T=55010T = 550 T=55T = 55

  2. For Car P:

    It travels at 25 m/s for T seconds then decelerates:

    Distance covered by P before deceleration: dP=25T+252(25T)=25T+252(25T)=25T+25(25T)2d_P = 25T + \frac{25}{2}(25 - T) = 25T + \frac{25}{2}(25 - T) = 25T + \frac{25(25 - T)}{2} Setting this equal to 800:

    Rearranging gives: 25T+25(25T)2=80025T + \frac{25(25 - T)}{2} = 800 By solving these equations, we find that the value of T is 9.

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