Post on 25-May-2015
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!