Kinematics Graphs

STARTER working in pairs describe each of the following graphs 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 2

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 - 5 -10 5 10 15 t (secs)

Kinematics Graphs

Without directly using the suvat equations WB1 Velocity time graph v (ms-1) 14 5 12 t (secs) Without directly using the suvat equations What can we work out from this graph?

Gradient 𝐚𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧= 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒊𝒎𝒆 WB1 Velocity time graph v (ms-1) 14 5 12 t (secs) Gradient 𝐚𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧= 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒊𝒎𝒆 𝐚= 𝟏𝟒−𝟓 𝟏𝟐 = 𝟑 𝟒 𝒎𝒔 −𝟐

Distance travelled 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭=𝐚𝐫𝐞𝐚 𝐮𝐧𝐝𝐞𝐫 𝐠𝐫𝐚𝐩𝐡 WB1 Velocity time graph v (ms-1) 14 5 12 t (secs) Distance travelled 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭=𝐚𝐫𝐞𝐚 𝐮𝐧𝐝𝐞𝐫 𝐠𝐫𝐚𝐩𝐡 𝐬= 𝟓 𝟏𝟐 + 𝟏 𝟐 𝟏𝟒−𝟓 𝟏𝟐 = 𝟏𝟎𝟓 𝒎

WB2 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 5s on the approach to the roadwork's, reducing the car’s speed from 28ms-1 to 22ms-1. The brakes are released and the car continues at a constant speed of 22ms-1 for a further 10 seconds. Sketch the speed time graph for this problem 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 WB2 28 v(ms-1) 22 Combining graphs and equations Answer: Straight line decrease in speed = constant deceleration 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 WB2 28 v(ms-1) 22 Combining graphs and equations Answer: a=(v-u)/t = (28 – 22) / 5 = 1.2 ms-2 this is the gradient of the v-t graph 5 15 t(s) b) Calculate the car’s deceleration in the first 5 seconds of the motion 𝑎= 𝑣−𝑢 𝑡 = 28 – 22 5 = 1.2 𝑚𝑠 −2 this is the gradient of the v-t graph

Combining graphs and equations Hint this journey is in two parts Use s=ut + ½ at2 for each part v(ms-1) 28 22 Combining graphs and equations 5 15 t(s) c) Find the total distance covered by the car during the 15 seconds Answer:for 1st part of journey s=22x5 + ½ x (-1.2)x 52 = 125 m + 2nd part of journey s = 22 x 10 = 220 makes 345 m in total This is the same as the area under the graph 1st part of journey 𝑠=22×5 + ½ × (−1.2)× 52 = 125 𝑚 + 2nd part of journey 𝑠 = 22×10 = 220 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 Combining graphs and equations 5 15 Answer: average speed = total distance / total time = 345 / 15 = 23 ms-1 t(s) d) What is the car’s average speed during the 15 seconds? average speed = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 = 345 15 = 23 𝑚𝑠 −1

A cyclist travels on a straight road over a 11 s period. WB3 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 t(s) 4 11 v(ms-1) 8 V

WB3 solution First – find their final velocity S U 8 V A 1 T 7 From t(s) 4 11 v(ms-1) 8 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 First – find their final velocity S U 8 V A 1 T 7 From t = 4 to t = 11 V = u + at V = 8 + 1 x 7 = 15 ms-1 Next: the total distance travelled is the area under the graph    

WB4 A man is jogging along a straight road at a constant speed of 4ms-1. He passes a friend with a bicycle who is standing at the side of the road and 20s 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. On the same diagram sketch the speed-time graphs of both cyclist and jogger Find the time that elapses before the cyclist meets the jogger

Cyclist and jogger answers (a) WB4 solution Let t = 0 at start Let t = T when they meet v (ms-1) Cyclist and jogger answers (a) 10 5 20 T t (secs)

Cyclist and jogger answers (b) WB4 solution v (ms-1) The jogger and cyclist will meet when they have gone the same distance 10 5 Distance travelled by jogger = 4T Cyclist and jogger answers (b) 20 24 T t (secs) 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 = 4T 24 + 12T – 288 = 4T T = 33 secs

WB5 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 Sketch, on the same diagram v-t graphs for both trains between X and Y [4] Given that XY = 975 m, find the value of T [8]

v (ms-1) M 40 N 30 20 10 T 25 𝑇 𝑌 t (secs) Two Trucks answers WB5 solution v (ms-1) M 40 N 30 Two Trucks answers 20 10 T 25 𝑇 𝑌 t (secs)

Train N distance travelled AREA = 25×30+ 1 2 𝑇 𝑌 −25 30 =975 WB5 solution v (ms-1) t secs 20 10 25 T 40 30 𝑇 𝑌 N M Train N distance travelled AREA = 25×30+ 1 2 𝑇 𝑌 −25 30 =975 Two Trucks answers Solves to 𝑇 𝑌 =43 Train M distance travelled AREA = 𝑇×40+ 1 2 43−𝑇 (40) =975 Solves to 𝑇= 33 4 =8.75

WB6 acceleration graph a(ms-2) - 4 4 1 2 t (secs) -8 3 -2 -6 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 Find the velocity of the particle when t = 2 At what time is the particle travelling in the negative direction?

Find the velocity of the particle when t = 2 WB6 acceleration graph solution Find the velocity of the particle when t = 2 At what time does the particle start travelling in the negative direction? 𝑎) 𝑣=𝑢 +𝑎𝑡 𝑣=8+ 4 2 =16 𝑚𝑠 −1 b) We want when the Velocity goes down to zero from 16 𝑚𝑠 −1 𝑣=𝑢 +𝑎𝑡 0=16+ −6 𝑡 𝑡= 16 6 =2.7 𝑠𝑒𝑐𝑠 So time after starting point is 2+2.7=4.7 𝑠𝑒𝑐𝑠

Velocity time graph v (ms-1) V U 3 T t (secs) WB7 proof A particle travels as shown on the graph Given initial velocity 𝒖=𝑼 𝒎𝒔 −𝟏 and constant acceleration 𝒂 =𝟏.𝟓 𝒎𝒔 −𝟐 And without directly using the suvat equations Show that the distance travelled D is given by 𝐷= 1 4 𝑇−3 4𝑈+3𝑇−9

Velocity time graph WB7 solution v (ms-1) V 3 t (secs) T U Gradient = 𝑉−𝑈 𝑇−3 =1.5 So 𝑽−𝑼= 3 2 (𝑇−3) Area under graph =𝑈 𝑇−3 + 1 2 𝑽−𝑼 (𝑇−3) =𝑈 𝑇−3 + 1 2 × 3 2 (𝑇−3)(𝑇−3) = 1 4 4𝑈 𝑇−3 +3 (𝑇−3)(𝑇−3) = 1 4 𝑇−3 4𝑈+3 𝑇−9)

WB8 Speed (ms-1) Time (s) 10 20 40 k 35 The speed-time graph above illustrates part of a car’s journey Find the magnitude of the car’s final deceleration 2ms-2 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. k=16

a = 0−10 5 =−2 so the magnitude of deceleration is 2 ms-2 WB8 solution Speed (ms-1) Time (s) 10 20 40 k 35 The speed-time graph above illustrates part of a car’s journey 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. a) v=u+at) a = 0−10 5 =−2 so the magnitude of deceleration is 2 ms-2 b) Area under graph = 20k + 10+𝑘 2 15 + 1 2 10×5 =540 20k + 75+7.5𝑘+25=540 𝑘=16 𝑚𝑠 −1

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:

One thing to improve is – 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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