KINEMATICS 3 KUS objectives BAT Draw accurate diagrams
KINEMATICS 3 KUS objectives BAT Draw accurate diagrams for distance-time and velocity time graphs BAT Find Area under a speed-time graph, gradient of a speed-time graph BAT Explore other graphs and links STARTER working in pairs describe each of the following graphs S displacement (distance travelled from a start point) U initial velocity V final velocity A constant acceleration T time spent on this section of a journey – a journey can be split into several sections, each of which has constant acceleration
STARTER DESCRIBE THIS GRAPH (i) Graph of a car journey Describe each of s u v a t s (m) 10 5 5 10 t (secs)
STARTER: DESCRIBE THIS GRAPH (ii) Graph of a car journey Describe each of s u v a t v (ms-1) 10 5 5 10 t (secs)
STARTER: DESCRIBE THIS GRAPH (iii) Graph of a car journey Describe each of s u v a t a(ms-2) 15 10 5 -10 5 10 15 t (secs)
WB 1 Velocity time graph v (ms-1) 14 5 12 t (secs)
WB 1 Velocity time graph v (ms-1) 14 5 12 t (secs)
WB 1 Velocity time graph v (ms-1) 14 5 12 t (secs)
WB 2 A car travels along a straight horizontal road on which there are some roadwork's with a speed restriction in force. The brakes are applied for 5 s on the approach to the roadwork's, reducing the car’s speed from 28 ms-1 to 22 ms-1. The brakes are released and the car continues at a constant speed of 22 ms-1 for a further 10 seconds. Sketch the speed time graph for this problem a) Explain how the speed-time graph shows when the brakes are applied the car undergoes a constant deceleration. b) Calculate the car’s deceleration in the first 5 seconds of the motion
Combining graphs and equations WB 2 28 v(ms-1) 22 5 15 t(s) a) Explain how the velocity-time graph shows when the brakes are applied the car undergoes a constant deceleration. a straight line decrease in ‘speed’ means the rate of change of velocity is constant
Combining graphs and equations WB 2 28 v(ms-1) 22 5 15 t(s) b) Calculate the car’s deceleration in the first 5 seconds of the motion
Combining graphs and equations Hint this journey is in two parts Use s=ut + ½ at 2 for each part v(ms-1) 28 22 5 15 t(s) c) Find the total distance covered by the car during the 15 seconds This makes 345 m in total. This is the same as the area under the graph
Combining graphs and equations 28 v(ms-1) 22 5 15 t(s) d) What is the car’s average speed during the 15 seconds?
WB 3 A cyclist travels on a straight road over a 11 s period. For the first four seconds they travel at constant speed of 8 ms-1 For the rest of the time they accelerate at a constant rate of 1 ms-2 Draw a Velocity time graph of their journey and work out the distance they have travelled v(ms-1) V 8 4 11 t(s)
WB 3 solution A cyclist travels on a straight road over a 11 s period. For the first four seconds they travel at constant speed of 8 ms-1 For the rest of the time they accelerate at a constant rate of 1 ms-2 Draw a Velocity time graph of their journey and work out the distance they have travelled v(ms-1) 8 4 First – find their final velocity From t = 4 to t = 11 S U 8 V A 1 T 7 V = u + at V = 8 + 1 x 7 = 15 ms-1 Next: the total distance travelled is the area under the graph 11 t(s)
WB 4 A man is jogging along a straight road at a constant speed of 4 ms-1. He passes a friend with a bicycle who is standing at the side of the road and 20 s later cycles to catch him up. The cyclist accelerates at a constant rate of 3 ms-2 until he reaches a speed of 12 ms-1. He then maintains a constant speed. a) On the same diagram sketch the speed-time graphs of both cyclist and jogger b) Find the time that elapses before the cyclist meets the jogger
WB 4 solution Let t = 0 at start Let t = T when they meet Cyclist and jogger answers (a) v (ms-1) 10 5 20 T t (secs)
WB 4 solution Cyclist and jogger answers (b) The jogger and cyclist will meet when they have gone the same distance v (ms-1) 10 5 Distance travelled by jogger = 4 T 20 24 Cyclist takes 4 secs to go from 0 to 12 ms-1 so reaches 12 ms-1 at time 24 secs Distance travelled by cyclist = ½ (24 - 20) x 12 + (T - 24) x 12 = 4 T Þ 24 + 12 T – 288 = 4 T Þ T = 33 secs T t (secs)
WB 5 Two trains M and N are moving in the same direction along parallel straight horizontal tracks. At time t=0, M overtakes N whilst they are travelling with speeds 40 ms-1 and 30 ms-1 respectively. Train M overtakes train N as they pass a point X at the side of the tracks. . After overtaking N, train M maintains its speed of 40 ms-1 for T seconds and then decelerates uniformly to rest next to a point Y on the tracks After being overtaken, train N maintains its speed of 30 ms-1 for 25 seconds and then decelerates uniformly, also coming to rest next to point Y The times taken by the trains to travel between X and Y are the same a) Sketch, on the same diagram v-t graphs for both trains between X and Y [4] b) Given that XY = 975 m, find the value of T [8]
WB 5 solution v (ms-1) 40 Two Trucks answers 30 M N 20 10 0 T 25 t (secs)
WB 5 solution v (ms 1) 40 M N 30 20 Two Trucks answers 10 0 T 25 t secs
WB 6 acceleration graph a(ms-2) 4 2 0 -2 - 4 -6 -8 1 2 3 4 t (secs) The acceleration-time graph models the motion of a particle. At time t = 0 the particle has a velocity of 8 ms-1 in the positive direction a) Find the velocity of the particle when t = 2 b) At what time is the particle travelling in the negative direction?
WB 6 acceleration graph solution a) Find the velocity of the particle when t = 2 b) At what time does the particle start travelling in the negative direction?
WB 7 proof Velocity time graph v (ms-1) V U 3 T t (secs)
WB 7 solution Velocity time graph v (ms-1) V U 3 T t (secs)
WB 8 k Speed (ms-1) 10 20 35 40 Time (s) The speed-time graph above illustrates part of a car’s journey a) Find the magnitude of the car’s final deceleration Given that the car travels 540 metres during this part of its journey (from t = 0 to t = 40) b) Find the value of k. 2 ms-2 k=16
WB 8 solution k The speed-time graph above illustrates part of a car’s journey a) Find the magnitude of the car’s final deceleration Speed (ms-1) 10 Given that the car travels 540 metres during this part of its journey (from t = 0 to t = 40) b) Find the value of k. 20 35 40 Time (s)
Summary Notes On a Displacement – time graph: Velocity is the gradient On a Velocity – time graph: Distance is the area under the graph On a Velocity – time graph: Acceleration is the gradient Also On an Acceleration – time graph: Velocity is the area under the graph Note:
KUS objectives BAT Draw accurate diagrams for distance-time and velocity time graphs BAT Find Area under a speed-time graph, gradient of a speed-time graph BAT Explore other graphs and links self-assess One thing learned is – One thing to improve is –
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