The diagram shows a velocity-time graph which models the motion of a car. The graph consists of four straight line segments. The car accelerates at a constant rate of 2 m s⁻² from rest to a speed of 20 m s⁻¹ over a period of T s. It then decelerates at a constant rate for 5 seconds before travelling at a constant speed of V m s⁻¹ for 27.5 s. The car then decelerates to rest at a constant rate over a period of 5 s.
- (a)Find T.[1]
- (b)Given that the distance travelled up to the point at which the car begins to move with constant speed is one third of the total distance travelled, find V.[4]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [1] (a) [2 = 20/T] → T = 10
- [4] (b) Distance travelled before constant speed = ½ × 10 × 20+ ½ × (20 + V) × 5
- [4] (b) May be implied if seen within total distance FT on T value from 4(a)
- [4] (b) Distance travelled after constant speed = 27.5V + ½ × 5V [= 30V]
- [4] (b) May be implied if seen within total distance
- [4] (b) ½ × 10 × 20 + ½ × (20 + V) × 5 = ⅓ [½ × 10 × 20 + ½ × (20 + V) × 5 + 27.5V + ½ × 5V]
- [4] (b) For attempting to use 1/2 or 1/3 correctly and for obtaining an equation for V which includes all parts of the journey. or ½ × 10 × 20 + ½ × (20 + V) × 5 = ½ [27.5V + ½ × 5V]
- [4] (b) V = 12
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