Kinematics of a Particle moving in a Straight Line

You can represent the motion of an object on a speed-time graph,

distance-time graph or an acceleration-time graph

2D

O

u

v

t

Initial velocity

Final velocity

Time taken

v - u

t

On a speed-time graph, the gradient of

a section is its acceleration!

v

u

t

On a speed-time graph, the Area beneath it is the

distance covered!

Kinematics of a Particle moving in a Straight Line

You can represent the motion of an object on a speed-time graph,

distance-time graph or an acceleration-time graph

Gradient of a speed-time graph = Acceleration over that period

Area under a speed-time graph = distance travelled during that period

2D

A car accelerates uniformly at 5ms-2 from rest for 20 seconds. It then travels at a constant speed for the next 40 seconds, then decelerates uniformly for the final 20 seconds until it is at rest again.a)Draw an acceleration-time graph for this informationb)Draw a distance-time graph for this information

20 40 60 80

5

Acceleration (ms-2)

0

-5

For now, we assume the rate of acceleration

jumps between different rates…

Time (s)

20 40 60 80 Time (s)

As the speed increases the curve gets steeper, but

with a constant speed the curve is straight. Finally the curve gets less steep

as deceleration takes place

Distance (m)

Kinematics of a Particle moving in a Straight Line

You can represent the motion of an object on a

speed-time graph, distance-time graph or an acceleration-time graph

Gradient of a speed-time graph = Acceleration over that

period

Area under a speed-time graph = distance travelled

during that period

2D

The diagram below shows a speed-time graph for the motion of a cyclist moving along a straight road for 12 seconds. For the first 8 seconds, she moves at a constant speed of 6ms-1. She then decelerates at a constant rate, stopping after a further 4 seconds. Find:a)The distance travelled by the cyclistb)The rate of deceleration of the cyclist

v(ms-1)

t(s)0

6

8 12

8

12

6

Sub in the appropriate values for the trapezium

above

Calculate

Kinematics of a Particle moving in a Straight Line

You can represent the motion of an object on a

speed-time graph, distance-time graph or an acceleration-time graph

Gradient of a speed-time graph = Acceleration over that

period

Area under a speed-time graph = distance travelled

during that period

2D

The diagram below shows a speed-time graph for the motion of a cyclist moving along a straight road for 12 seconds. For the first 8 seconds, she moves at a constant speed of 6ms-1. She then decelerates at a constant rate, stopping after a further 4 seconds. Find:a)The distance travelled by the cyclist – 60mb)The rate of deceleration of the cyclist

v(ms-1)

t(s)0

6

8 124

-6

Sub in the appropriate values for the trapezium

above

Calculate

Kinematics of a Particle moving in a Straight Line

You can represent the motion of an object on a

speed-time graph, distance-time graph or an acceleration-time graph

Gradient of a speed-time graph = Acceleration over that

period

Area under a speed-time graph = distance travelled

during that period

2D

A particle moves along a straight line. It accelerates uniformly from rest to a speed of 8ms-1 in T seconds. The particle then travels at a constant speed for 5T seconds. It then decelerates to rest uniformly over the next 40 seconds.a)Sketch a speed-time graph for this motionb)Given that the particle travels 600m, find the value of Tc)Sketch an acceleration-time graph for this motion

v(ms-1)

t(s)0

8

T 5T 40

5T

8

6T + 40

Sub in values

Simplify fraction

Divide by 8

Subtract 20

Divide by 5.5

Kinematics of a Particle moving in a Straight Line

You can represent the motion of an object on a

speed-time graph, distance-time graph or an acceleration-time graph

Gradient of a speed-time graph = Acceleration over that

period

Area under a speed-time graph = distance travelled

during that period

2D

A particle moves along a straight line. It accelerates uniformly from rest to a speed of 8ms-1 in T seconds. The particle then travels at a constant speed for 5T seconds. It then decelerates to rest uniformly over the next 40 seconds.a)Sketch a speed-time graph for this motionb)Given that the particle travels 600m, find the value of T – 10 secondsc)Sketch an acceleration-time graph for this motionv(ms-1)

t(s)0

8

T 5T 405010

First section Last section

t(s)

a(ms-2)

20 40 60 80 100

0.8

-0.2

Kinematics of a Particle moving in a Straight Line

You can represent the motion of an object on a

speed-time graph, distance-time graph or an acceleration-time graph

Gradient of a speed-time graph = Acceleration over that

period

Area under a speed-time graph = distance travelled

during that period

2D

A car C is moving along a straight road with constant speed 17.5ms-1. At time t = 0, C passes a lay-by. Also at time t = 0, a second car, D, leaves the lay-by. Car D accelerates from rest to a speed of 20ms-1 in 15 seconds and then maintains this speed. Car D passes car C at a road sign.a)Sketch a speed-time graph to show the motion of both carsb)Calculate the distance between the lay-by and the road sign

v(ms-1)

t(s)0

20

17.5

15

C

DAt the road sign, the cars have

covered the same distance in the same time

We need to set up simultaneous equations using s and t…

Let us call the time when the areas are equal ‘T’

T

17.5

T

T - 15

20

Sub in values

Sub in values

Simplify fraction

Multiply bracket

Kinematics of a Particle moving in a Straight Line

You can represent the motion of an object on a

speed-time graph, distance-time graph or an acceleration-time graph

Gradient of a speed-time graph = Acceleration over that

period

Area under a speed-time graph = distance travelled

during that period

2D

A car C is moving along a straight road with constant speed 17.5ms-1. At time t = 0, C passes a lay-by. Also at time t = 0, a second car, D, leaves the lay-by. Car D accelerates from rest to a speed of 20ms-1 in 15 seconds and then maintains this speed. Car D passes car C at a road sign.a)Sketch a speed-time graph to show the motion of both carsb)Calculate the distance between the lay-by and the road sign

v(ms-1)

t(s)0

20

17.5

15

C

DAt the road sign, the cars have

covered the same distance in the same time

We need to set up simultaneous equations using s and t…

Let us call the time when the areas are equal ‘T’

T

Subtract 17.5T

Add 150Divide by 2.5

Sub in T

Calculate!

Set these equations equal to each other!