Velocity-Time Graphs & Distance FINAL

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Velocity-Time Graphs & Distance

Velocity-Time Graphs & Distance: 

Velocity-Time Graphs & Distance To be able to calculate distance from a velocity-time graph

Acceleration: 

Acceleration Acceleration is the change in velocity (ΔV) ÷ time (t) Acceleration (a) Change In Velocity (ΔV) Time (t)

Velocity-Time Graphs: 

Velocity-Time Graphs Remember, the velocity of an object is its speed in a particular direction (this means that two cars travelling at the same speed, but in opposite directions, have different velocities) When an object is moving with a constant velocity, the line on the graph is horizontal When an object is moving with a constant acceleration, the line on the graph is straight, but sloped The steeper the line, the greater the acceleration of the object

Acceleration: 

Acceleration Acceleration is represented on a velocity-time graph by the gradient of the line (change in velocity ÷ time) What is the acceleration - represented by the sloping line? Change in velocity from 0m/s to 8m/s = 8m/s Time of 4 seconds for change in velocity 8 ÷ 4 = 2m/s 2

Distance: 

Distance The area under the line on a velocity-time graph represents the distance travelled What distance was covered on the above graph?

Distance: 

Distance To find the distance we need to calculate the area of the light blue and dark blue regions For rectangle areas use the formula base x height 6s x 8m/s = 48m For triangle areas use the formula ½ x base x height ½ x 4s x 8m/s = 16m Distance = 48 + 16 = 64m

Graphing: 

Graphing Let us revise and check our knowledge about graphs with the help of few questions.

Question 1: 

Question 1 Match A, B, C and D to the following descriptions: - Accelerated motion throughout Zero acceleration Accelerated motion, then decelerated motion Deceleration Which line represents the furthest distance? Which line represents the least distance?

Answer 1: 

Answer 1 Match A, B, C and D to the following descriptions: - Accelerated motion throughout – A ( 1/2 x 20s x 8m/s = 80m) Zero acceleration – C (20s x 8m/s = 160m) Accelerated motion, then decelerated motion – D ( 1/2 x 20s x 6m/s = 60m) Deceleration – B ( 1/2 x 20s x 4m/s = 40m) Which line represents the furthest distance? – C Which line represents the least distance? – B

Question 2: 

Question 2 Describe the motion of the cyclist Work out the initial acceleration Work out the distance travelled by the cyclist in the first 40 seconds

Answer 2: 

Answer 2 Describe the motion of the cyclist – accelerates at constant rate for 40 seconds, then decelerates for the next 20 seconds to a standstill Work out the initial acceleration – 0.2m/s 2 (8m/s ÷ 40s) Work out the distance travelled by the cyclist in the first 40 seconds – ½ x 40s x 8m/s = 160m

Question 3: 

Question 3 In a motorcycle test the speed from rest was recorded at intervals Time (s) 0 5 10 15 20 25 30 Velocity (m/s) 0 10 20 30 40 40 40 Plot a velocity-time graph of these results What was the initial acceleration? How far did it travel in the first 20 seconds? How far did it travel in the next 10 seconds?

Answer 3: 

Answer 3 What was the initial acceleration – 2m/s 2 (40m/s ÷ 20s) How far did it travel in the first 20 seconds – ½ x 20s x 40m/s = 400m How far did it travel in the next 10 seconds – 40 x 10 = 400m

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THANK YOU GAURAV AGGARWAL XI-B ROLL NO.-13