Figure 1 Figure 1 shows the speed-time graph for the journey of a car moving in a long queue of traffic on a straight road. At time t=0, the car is at rest at the point A. The car then accelerates uniformly for 5 seconds until it reaches a speed of 5ms^-1 For the next 15 seconds the car travels at a constant speed of 5ms^-1 The car then decelerates uniformly until it comes to rest at the point B. The total journey time is 30 seconds. (a) Find the distance AB. (b) Sketch a distance-time graph for the journey of the car from A to B. __ (3) (3)

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Figure 1 Figure 1 shows the speed-time graph for the journey of a car moving in a long queue of traffic on a straight road. At time t=0, the car is at rest at the point A. The car then accelerates uniformly for 5 seconds until it reaches a speed of 5ms^-1 For the next 15 seconds the car travels at a constant speed of 5ms^-1 The car then decelerates uniformly until it comes to rest at the point B. The total journey time is 30 seconds. (a) Find the distance AB. (b) Sketch a distance-time graph for the journey of the car from A to B. __ (3) (3)

Answer 1

### (a) The distance \(AB\) is \(112.5 \, \mathrm{m}\). ### (b) Sketch the distance-time graph as described in Step6.

Answer 2

## Step1: Analyze the Speed-Time Graph ### The graph shows three distinct phases: acceleration, constant speed, and deceleration. We need to calculate the distance covered in each phase. ## Step2: Calculate Distance During Acceleration ### The car accelerates uniformly from rest to \(5 \, \mathrm{m/s}\) in 5 seconds. The distance covered during acceleration can be found using the formula for the area of a triangle: \[\text{Distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \, \mathrm{s} \times 5 \, \mathrm{m/s} = 12.5 \, \mathrm{m}\] ## Step3: Calculate Distance During Constant Speed ### The car travels at a constant speed of \(5 \, \mathrm{m/s}\) for 15 seconds. The distance covered during this phase is: \[\text{Distance} = \text{speed} \times \text{time} = 5 \, \mathrm{m/s} \times 15 \, \mathrm{s} = 75 \, \mathrm{m}\] ## Step4: Calculate Distance During Deceleration ### The car decelerates uniformly to rest in the remaining 10 seconds. The distance covered during deceleration can be found using the formula for the area of a triangle: \[\text{Distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \, \mathrm{s} \times 5 \, \mathrm{m/s} = 25 \, \mathrm{m}\] ## Step5: Sum the Distances ### The total distance \(AB\) is the sum of the distances covered in each phase: \[\text{Total Distance} = 12.5 \, \mathrm{m} + 75 \, \mathrm{m} + 25 \, \mathrm{m} = 112.5 \, \mathrm{m}\] ## Step6: Sketch the Distance-Time Graph ### The distance-time graph will have three segments: 1. A curve representing acceleration from \(0 \, \mathrm{m}\) to \(12.5 \, \mathrm{m}\) over 5 seconds. 2. A straight line representing constant speed from \(12.5 \, \mathrm{m}\) to \(87.5 \, \mathrm{m}\) over the next 15 seconds. 3. A curve representing deceleration from \(87.5 \, \mathrm{m}\) to \(112.5 \, \mathrm{m}\) over the final 10 seconds. #

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