1. Velocity - time graph 1. The velocity – time graph shows the motion of a particle for one minute. Calculate each of the following. (a) The acceleration in the first 10s. (b) The total distance travelled. (c) The total displacement. (d) The average speed. (e) The average velocity. (a) a = 6/10 = 0.6 ms -2. (b) D = ½  10  6+20  6+ ½  15  6 + ½  15  6 = 240m (c) s = 30 + 120 + 45 – 45 = 150m (d) = 240  60 = 4 ms -1 (e) = 150  60 = 2.5 ms -1. 30 120 45

2. Velocity - time graph 2. A car leaves a point O and accelerates uniformly from rest to a speed of 6 ms -1 in 4 seconds. It then maintains a steady speed for a further 10 seconds, after which it slows down to a halt in 2 seconds. (a) Sketch a speed–time graph for the car’s journey. Speed ms -1 6 t=4 t=14 t=16 time (t)

3. Velocity - time graph (b) Find the acceleration during the initial and final stages of the journey. Speed ms -1 6 t=4 t=14 t=16 time (t) The initial stage: Acceleration = gradient = 6  4 = 1.5 ms -2 The final stage: Acceleration = gradient = -6  2 = -3 ms -2

4. Velocity - time graph (c) Find the total distance travelled by the car. Speed ms -1 6 t=4 t=14 t=16 time (t) Distance travelled = Total area: Area = ½  6  4 + 6  10 + ½  6  2 = 12 + 60 + 6 = 78 m 12 60 6

5. 3. Two cars set off on a journey. The first car leaves at time t = 0, where t is measured in seconds. It accelerates uniformly until it reaches a speed of 8 m s –1 at t = 3 s. It then maintains a constant velocity. The second car leaves from the same point at t = 3 and travels with constant speed 16 m s –1. (a) On the same axes sketch speed–time graphs for the motion of the two cars. time (t) (seconds) (m s –1 ) speed first car second car

6. (b) Find t when the two cars meet. time (t) (seconds) (m s –1 ) speed first car second car Suppose the cars meet after t seconds: First car: s = ½  8  3+ 8(t – 3) Second car: s = 16(t – 3) 16(t – 3) = ½  8  3+ 8(t – 3)  16t – 48 = 12 + 8t - 24 8t = 36  t = 4.5 s

7. (c) How far are they from the start at this time ? Distance travelled by the first car = So, they meet 24 m from the start. 16( t – 3) = 16( 4.5 – 3) = 24m d) Sketch a displacement–time graph for the journey of the second car. 0 3 time (t) Displacement m

8. 4. The graph shows the motion of a bungee jumper. (a) State two non-zero times at which the velocity of the jumper is zero. (b) Find the total distance travelled and the final displacement. (a) t = 4 s and t = 5.6 s (b) D = ½  4  25 + ½  1.6  10 = 58m s = ½  4  25 - ½  1.6  10 = 42m below starting position. 50 8