314271156-distance-time-and-velocity-time-graphs-edit.pdf

Distance Time and Velocity Time Graphs Preliminaries Distance time graphs are so called because of the simple fact that the distance an object is plotted against the time it takes to complete the journey. Velocity graphs are similar in that the velocity/speed of an object is plotted against the time it takes to complete the journey. Formula

Speed 

Distance time

Average Speed 

Acceleration 

Total Distance Total time

Speed time

And Deceleration  

Speed time

When attempting to solve problems involving distance time graphs care should be taken with the following 1. Pay attention to the scales used on the x and y axes and the units used 2. Read the question carefully 3. Note that while distance time and velocity time graphs may have the same shape the meaning given to the each line is different.

Deceleration and acceleration are vector quantities and are opposite to each other. Acceleration is given a positive value and deceleration a negative value

Study the diagrams below and note the differences in both types of graphs

Distance

The Distance time Graph

The Velocity time graph

Velocity/Speed

Constant Speed/cruising speed

Using Distance time Graphs Practice Question 1: Consider the graph below

The graph shows the movement of a car from point A to point D, D is 500m from A note that there are two slant lines AB and CD. The slant lines indicate that the car is moving. The Flat line BC indicates that the car has stopped or is at rest. Discussion questions 1. Calculate the speed of the car during the first 10 seconds 2. For how long did the car stop? 3. What is the speed of the car on its journey from C to D 4. On which part of the journey did the car travel faster? 5. What is the average speed of the car for the whole journey 6. What is the average speed of the car for the time it was moving?

Answers

1. We use the formula Speed 

Distance , please notice that you are actually calculating time

the gradient of the line AB using the formula

S

Rise . This gives us Run

200m  20m / s note that this can be written as 20ms 1 10 s

2. Notice that one small box on the x axis is used to represent 5 seconds. This means that the car stopped for 20 seconds 3. Using Speed 

S

Distance same as finding the gradient of the line, we have time

300m  10m / s note that this can be written as 10ms 1 30 s

4. Comparing the speeds for the two sections we can see that that car travelled faster on the AB leg of the journey. You could have seen that from the graph as well where line AB is steeper than line CD. The steeper the line is the bigger the gradient is. 5. Using the formula Average Speed 

Total Distance 500m we have S   8.3m / s Total time 60 s

6. Note that the car wasn’t moving for 20 seconds which meant it was moving for 40 seconds, using the same formula Speed 

Distance 500m we have S   12.5m / s time 40 s

Practice Question 2

Solutions: Please take a moment to look carefully at the scales/axes and see how much one box is representing before you go any further (i)

(ii) (iii)

This requires us to calculate the gradient of the line, this gives us Rise 12   6km / h Run 2 The athlete was resting at this time. The athlete actually rested for an hour and a half The average speed for the return journey requires the same idea as before. We need Rise 12 the gradient of the line   8km / h Run 1.5

Practice Question 3

Solution

(i)

It was there for 20 minutes, resting, waiting, it stopped, whatever it did

(ii)

The average speed here is the same as finding the gradient of the line Average Speed 

Total Distance  40   100    Total time  60 

 60  100     150km / h  40 

(iii)

Because the time is in minutes we have to write it as a fraction of an hour

How long will the train take to travel 50km at 60km/h? we use the formula 50 Distance time , , it takes 50 minutes Speed  50 time time  min 60 60 

(iv)

The answer to part (iv) is on the graph//red line

Velocity time graphs Practice Question: Consider the graph below: Notice that the y axis says velocity

Discussion questions 1. Calculate the acceleration of the athlete during the first 4 seconds 2. What was the athlete doing during between the 4th and 14th second 3. Calculate the deceleration of in the final stage of the race 4. How far has the athlete moved in the first 4 seconds? 5. What is the total distance travelled by the athlete Solutions 1. The slant line/line segment from 0 – 4 shows the athlete speeding up, we find the acceleration by finding the gradient of the line Rise 8m   2m / s this can be written as 2ms 2 Run 4 s

2. The athlete is travelling at a constant speed of 8m / s

3. This part is the same as in number 1, we find the gradient but assign it a negative value. Rise 8m   8m / s 2 this can be written as 8ms 2 Run 1s

4. To calculate the distance travelled we need to calculate the area of the geometric shape formed. Notice that during the first 4 seconds we have a triangle We will therefore find the area of the triangle formed which gives

8 4  16m 2 So the athlete travelled 16 meters in the first 4 seconds

5. To find the total distance travelled we must find the area of the shape made for the race. This is a trapezium so we use the trapezium rule We find the area of the trapezium

1 ha  b 2 1 A   8 16  11  108m 2 A

Practice Question 2

Solution (i) (ii)

40 8  m / s2 15 3 To find the distance travelled between the period t = 15 and t = 35 we have to find the area of the shape formed as outlined below

We need to find the gradient of the first line segment, this gives us

We find the area of the trapezium

1 ha  b 2 1 A   20  40  50  2 A  900m A

Practice Questions from CXC/CSEC