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SITEMAP   School-college Physics Notes: Forces & motion Part 1.3 Distance-time graphs

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Forces and Motion 1.3 Drawing and interpreting distance - time graphs and calculations - graphical problem solving

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INDEX for physics notes on speed calculations and constructing and interpreting distance-time graphs


This page contains online questions only. Jot down your answers and check them against the worked out answers at the end of the page

1.3 Drawing and interpreting distance - time graphs and calculations - problem solving

The gradient (slope) at any point on the graph gives you the speed at that point.

Since speed = distance time, then for graphical work

speed = (change in vertical y axis for distance) (change in horizontal x axis for time)

speed = ∆d / ∆t

The steeper the gradient the greater the speed

(1)

Distance - time graphs - acceleration - speeding up

Graph  curves upwards, showing increasing speed/velocity with time (acceleration), for each incremental time unit (e.g. minute or second) there is an ever increasing (larger) distance (∆d) covered in the same time (∆t).

speed = ∆d/∆t = the gradient of graph (∆d/∆t) is continuously increasing.

 

(2)

Distance - time graphs - motionless - stationary

Graph  is flat/horizontal, indicating zero speed/velocity, there is no increase in distance with time, object has stopped moving.

 

(3)

Distance - time graphs - deceleration - slowing down

Graph shows the curve is levelling off, steadily decreasing speed/velocity with time (deceleration), for each incremental time unit (e.g. minute or second) there is an ever decreasing (smaller) distance covered in the same time.

 

(4)

Distance - time graphs - constant speed

Graph is linear, showing constant speed/velocity, the distance covered in any equal time increment is the same.

 

The four graphs above illustrate of what you might see at any point on a distance-time graph, BUT, in reality, any distance-time graph will be more complicated than any of these specific graphs. So, you find a graph with all four types of gradient in just one distance-time graph - see the graph based questions below.

 

Examples of distance-time graph calculation questions

Moving on from the stylised 'types of graph' to full numerical graphs.

distance - time graphs interpreting measuring gradient speed calculations gcse physics igcse international GCSE physics

Graph (1) The graph lines for objects A and B are linear, indicating constant speed.

Using a convenient triangle to calculate the constant speeds of A and B

Speed of A = 500 / 10 = 50 m/s   and   speed of B = 300 / 25 =  12 m/s

Note the steeper the gradient, the greater the speed.

Graph (2) The graph line is horizontal, no change in distance, so object C is stationary at a distance of 300 m.

Graph (3) Lets call the object DE:

Its movement is first in a forward direction, represented by line D at a constant speed of 400 / 5 = 80 m/s.

Its second movement is in the opposite direction, line E, at a constant speed of 400 / 10 = 40 m/s.

(Strictly speaking, the line E represents a speed of -40 m/s, because of the reversed direction.)

Note that object DE has travelled a total distance of 800 m, BUT, its displacement is zero!

 

You get the speed from the gradient at any given point on the graph, which is easy if the graph at that point is linear (constant speed, like the three graphs above).

For example look at the linear portion between the two purple 'blobs' on the right-hand graph - that's a linear section and easy to measure the gradient (speed).

If the graph is curved you will need to draw a tangent to the curve at the point where you want to know the speed.

For example I've drawn a tangent at ~56 seconds and converted it to a purple triangle so you can measure the gradient - see Q2.1 (c) below.


Questions based on distance-time graphs

Q2.1 The graph below shows part of a car journey.

Distance - time graph for a short car journey

Interpretation question using speed = distance / time

(a) Including calculating the average speed, describe and explain the motion of the car between:

(i) 0 and 20 seconds?

(ii) 20 and 40 seconds?

(iii) 40 to 82 seconds?

(iv) 82 to 90 seconds?

(b) What is the average speed of the car while it is moving?

(c) How can you find the specific speed of the car  at any point on the graph?

(d) What is the speed of car at 56 seconds?

Worked out ANSWERS to distance-time graph questions

 

Q2.2 The graph below summarises part of the journey of a train (somewhat delayed at some point!).

Interpretation question using speed = distance / time and giving your answers in km/hour and m/s

(a) What is the average speed between 1300 and 1400 hours?

(b) What is the average speed between 1530 and 1800 hours?

(c) How long was the train stopped for?

(d) What was the average speed for the whole journey?

Worked out ANSWERS to distance-time graph questions


Velocity - Time graphs and acceleration are dealt with in detail in section 2

INDEX physics notes: Speed calculations and distance-time graphs


Keywords, phrases and learning objectives for speed, motion and graphs

Be able to draw a distance-time graph from given data of a moving object.

Be able to interpret a distance - time graphs and associated calculations reading data from the graph.


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INDEX physics notes: Speed calculations and distance-time graphs

Worked out ANSWERS to the distance-time graph questions

Q2.1 The graph below shows part of a car journey.

Distance - time graph for a short car journey

Interpretation question using speed = distance / time

(a) Including calculating the average speed, describe and explain the motion of the car between:

(i) 0 and 20 seconds?

speed increasing (acceleration) because the gradient is steadily increasing

time interval = 20 - 0 = 20 s, distance covered = 200 - 0 = 200 m

gradient = average speed = 200 / 20 = 10 m/s

(ii) 20 and 40 seconds?

gradient is constant, but NOT zero, car moving at steady constant speed

time interval = 40 - 20 = 20 s, distance covered = 600 - 200 = 400 m

gradient = constant speed = 400 / 20 = 20 m/s

(iii) 40 to 82 seconds?

gradient decreasing (deceleration), gradient steadily decreasing

time interval = 82 - 40 = 42 s, distance covered = 900 - 600 = 300 m

gradient = average speed = 300 / 42 = 7.14 m/s ( 1 dp, 3 sf)

(iv) 82 to 90 seconds?

At 82 seconds the gradient is zero, graph line is flat, so car isn't moving - stationary from 82 to 90 s.

(b) What is the average speed of the car while it is moving?

total time of journey (while moving) is 82 s.

Total distance covered until it stops = 900 m

average speed for journey = 900 / 82 = 10.8 m/s (1 dp, 3 sf)

(c) How can you find the specific speed of the car  at any point on the graph?

You can find the speed at any point on the graph by drawing a tangent at that point on the graph.

No tangent is needed if the graph line is linear, but you need to draw a tangent for any point on the curved sections of a speed/velocity-time graph.

(d) what is the speed of car at 56 seconds?

The same graph as above but tangent drawn on at 56 seconds.

You convert the purple tangent line into a triangle so you read of the distance travelled (vertical y axis) and the time taken (horizontal x axis) and then calculate the gradient from the two dimensions of the triangle, which is the gradient at the point you are interested in.

x-axis: time Δt = 80 - 16 = 64 s (easy),

y-axis: distance Δd = 1000 - 480 (watch the scale!) = 520 m

speed = distance / time taken = Δd/Δt = 520 / 64 = 8.1 m/s (to 2 s.f.)

 

Q2.2 The graph below summarises part of the journey of a train (somewhat delayed at some point!).

Interpretation question using speed = distance / time and giving your answers in km/hour and m/s

(a) What is the average speed between 1300 and 1400 hours?

time interval = 1 hour, distance travelled = 100 - 0 = 100 km

average speed = 100/1 = 100 km/hour

1 km = 1000 m, 1 hour = 3600 s, average speed = 100 x 1000 / 3600 = 27.8 m/s (1 dp, 3 sf)

(b) What is the average speed between 1530 and 1800 hours?

time interval = 2.5 hours, distance travelled = 250 - 100 = 150 km

average speed = 150/2.5 = 60 km/hour

1 km = 1000 m, 1 hour = 3600 s, average speed = 60 x 1000 / 3600 = 16.7 m/s (1 dp, 3 sf)

(c) How long was the train stopped for?

from 1400 to 1530 hours, 1.5 hours.

(d) What was the average speed for the whole journey?

total time from 1300 to 1800 hours = 5 hours, total distance travelled = 250 km

average speed = 250/5 = 50 km/hour

1 km = 1000 m, 1 hour = 3600 s, average speed = 50 x 1000 / 3600 = 13.9 m/s (1 dp, 3 sf)

 

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INDEX physics notes: Speed calculations and distance-time graphs

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