SPEED/TIME GRAPHS. Consider the following Speed/Time graphs: v (ms –1 ) 15 6 t (s) This shows:A...

SPEED/TIME GRAPHS

Consider the following Speed/Time graphs:

v (ms–1)

15

6t (s)

This shows: A body moving at 15 ms–1,i.e. zero acceleration,shown by a gradient of 0.In 6 seconds, it will move 90m.

(Given by the area)

This shows:v (ms–1)

20

4t (s)

A body accelerating fromrest to 20 ms–1, in 4 seconds,i.e. at 5 ms–2.

(Given by the gradient)

Also, the distance moved will be equal to the area of thetriangle underneath the line, i.e. 2 × 20 = 40 m.

Example 1: A tube train moving between two stations accelerates uniformly from rest to 20 ms–1. It then travels with constant speed for twice as long as it accelerated. The total distance covered in these two stages was 800 m. The tube train then retards at a constant rate back to rest. Given the total time for the journey was 56 seconds, find the total distance covered and the deceleration in the final stage.v (ms–1)

t (s)

20

t1 t22t1

Using: Area under the graph = distance:In the 1st two stages:

10t1 + 40t1 = 800t1 =

Considering the total time:3t1 + t2 = 56

t2 = Total distance = 800 + 4 × 20 = 880 m

Using: Gradient = acceleration:

In the 3rd stage: acceleration = –20 8 = –2.5 deceleration = 2.5 ms–2

16

8

Example 2: A train accelerates at ⅔ ms–2 from rest to a speed V ms–1. It then moves with a constant speed V ms–1 for 45 seconds and then decelerates back to rest, for half of the time that it accelerated. The whole journey is 1.35 km. Find the value of V.

2T T

V

v (ms–1)

t (s)45

Using:

Using: Area under the graph = distance:

In the 1st stage:

= V2T

23

TV + 45V + 12 TV = 1350

T = 3V 4

45V +3V 4

2+ 3V

8

2= 1350

This leads to: V2 + 40V – 1200 = 0(V – 20)(V + 60) = 0 Hence: V = 20

… (1)

sub in (1)

Gradient = acceleration:

Summary of key points:

This PowerPoint produced by R Collins; © ZigZag Education 2008-2010

On a Speed/Time graph:

The gradient = acceleration*

The area under the graph = distance travelled*