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CH-14-001 · Velocity–time graph: distance and average speed

The velocity–time graph shows a vehicle moving in a straight line. It accelerates uniformly from rest to 18 m/s in 6 s, travels at this velocity for 25 s, and then slows uniformly to rest in 5 s.
018063136Time (s)Velocity (m/s)

(a) Calculate the total distance travelled and the average speed for the whole journey. Give the speed to 2 decimal places.

(b) Find the magnitude of the deceleration.

Part (a) Velocity–Time Graphs: Area (Distance), DuSTy & Gradient Booklet p. 54 & p. 58
Method
1) Distance = area under V-T graph (triangle + rectangle + triangle)
2) Average speed = Total Distance ÷ Total Time (DuSTy)
3) Magnitude of deceleration = absolute value of the final gradient
Worked Example (CH-14-001 · Part (a))
Calculate the total distance travelled and the average speed for the whole journey. Give the speed to 2 decimal places.
Booklet p. 58: Area under V–T graph = Distance (Area 1◯+Area 2◯+Area 3◯\text{Area }\textcircled{1} + \text{Area }\textcircled{2} + \text{Area }\textcircled{3}) · Gradient = Acceleration (y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1})
1) First triangular area = 54.0 m.
2) Rectangular area = 450 m and final triangular area = 45.0 m.
3) Total distance 549.00 m.
4) Divide by total time 36 s.
5) 15.25 m/s.
549.00 m; 15.25 m/s.
Part (b) Velocity–Time Graphs: Area (Distance), DuSTy & Gradient Booklet p. 54 & p. 58
Method
1) Distance = area under V-T graph (triangle + rectangle + triangle)
2) Average speed = Total Distance ÷ Total Time (DuSTy)
3) Magnitude of deceleration = absolute value of the final gradient
Worked Example (CH-14-001 · Part (b))
Find the magnitude of the deceleration.
Booklet p. 58: Area under V–T graph = Distance (Area 1◯+Area 2◯+Area 3◯\text{Area }\textcircled{1} + \text{Area }\textcircled{2} + \text{Area }\textcircled{3}) · Gradient = Acceleration (y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1})
1) 18 ÷ 5 = 3.60 m/s².
3.60 m/s²