SPEED/TIME GRAPHS

Consider the following Speed/Time graphs: v (ms –1 ) 15 6 t (s) This shows:A body moving at 15 ms –1, i.e. zero acceleration, shown by a gradient of 0. In 6 seconds, it will move90m. (Given by the area) This shows: v (ms –1 ) 20 4 t (s) A body accelerating from rest to 20 ms –1, in 4 seconds, i.e. at 5 ms –2. (Given by the gradient) Also, the distance moved will be equal to the area of the triangle underneath the line, i.e. 2 × 20= 40 m.

Example 1: A tube train moving between two stations accelerates uniformly from rest to 20 ms – 1. It then travels with constant speed for twice as long as it accelerated. The total distance covered in these two stages was 800 m. The tube train then retards at a constant rate back to rest. Given the total time for the journey was 56 seconds, find the total distance covered and the deceleration in the final stage. v (ms –1 ) t (s) 20 t1t1 t2t2 2t12t1 Using: Area under the graph = distance: In the 1 st two stages: 10t t 1 = 800 t 1 = Considering the total time: 3t 1 + t 2 = 56 t 2 = Total distance = × 20 = 880 m Using: Gradient = acceleration: In the 3 rd stage: acceleration = –20 8 = –2.5 deceleration = 2.5 ms –2 16 8

Example 2:A train accelerates at ⅔ ms – 2 from rest to a speed V ms – 1. It then moves with a constant speed V ms – 1 for 45 seconds and then decelerates back to rest, for half of the time that it accelerated. The whole journey is 1.35 km. Find the value of V. 2TT V v (ms –1 ) t (s) 45 Using: Using: Area under the graph = distance: In the 1 st stage: = V 2T 2323 TV +45V TV = 1350 T = 3V 4 45V + 3V V 8 2 = 1350 This leads to:V V – 1200 = 0 (V – 20)(V + 60) = 0 Hence: V = 20 … (1) sub in (1) Gradient = acceleration:

Summary of key points: This PowerPoint produced by R Collins; © ZigZag Education On a Speed/Time graph: The gradient = acceleration * The area under the graph = distance travelled *