Distance-Time Graph | Speed-Time Graph | 9th Physics | Example 2.6 | Example 2.7,2.8,2.9 | M Shaban
GRAPHICAL ANALYSIS OF MOTION
Graph is a pictorial way of presenting information about the relation between various quantities. The quantities between which a graph is plotted are called the variables. One of the quantities is called the independent quantity and the other quantity, the value of which varies with the independent quantity is called the dependent quantity.
For example y=5x.
Here y is dependent and x is independent.
DISTANCE-TIME GRAPH
It is useful to represent the motion of objects using graphs. The terms distance and displacement are used interchangeably when the motion is in a straight line. Similarly if the motion is in a straight line then speed and velocity are also used interchangeably. In a distance-time graph, time is taken along horizontal axis while vertical axis shows the distance covered by the object.
OBJECT AT REST
In the graph shown in figure 2.18, the distance moved by the object with time is zero. That is, the object is at rest. Thus a horizontal line parallel to time axis on a distance-time graph shows that speed of the object is zero.
OBJECT MOVING WITH CONSTANT SPEED
If an object covers equal distance in equal intervals of time, it is called constant speed.
The speed during the first five seconds.
Average speed of the car.
Speed during the last 5 seconds.
Solution:
(a) Total distance travelled = 40m
(b) Distance travelled during first 5s is 35 m
Speed=distance/time=35m/5s=7ms^(-1)
(c)
Average speed=Distance/time=40m/10s=4ms^(-1)
(d) Distance moved during the last 5s = 5 m. So
Speed=5m/5s=1ms^(-1)
Required answer.
SPEED-TIME GRAPH
In a speed-time graph, time is taken along x-axis and speed is taken along y-axis.
OBJECT MOVING WITH CONSTANT SPEED
When the speed of an object is constant (4 ms-1 ) with time, then the speed-time graph will be a horizontal line parallel to time-axis along x-axis as shown in figure 2.22. In other words, a straight line parallel to time axis represents constant speed of the object.
OBJECT MOVING WITH UNIFORMLY CHANGING SPEED (uniform acceleration)
Let the speed of an object be changing uniformly. In such a case speed is changing at constant rate. Thus its speed-time graph would be a straight line such as shown in figure 2.23. A straight line means that the object is moving with uniform acceleration. Slope of the line gives the magnitude of its acceleration.
Example 2.7
Find the acceleration from speed-time graph shown in figure 2.23.
Solution:
On the graph in figure 2.23, point A gives speed of the object as 2 ms-1 after 5 s and point B gives speed of the object as 4 ms-1 after 10 s. So acceleration is given as
a ⃑=(change in velocity)/(change in time)=(4ms^(-1)-2ms^(-1))/(10s-5s)=(2ms^(-1))/5s=0.4ms^(-2)
Speed-time graph in figure 2.23 gives acceleration of the object as 0.4 ms-2.
Example 2.8
Find the acceleration from speed-time graph shown in figure 2.24.
Solution:
At ti =5s is vi = 4ms-1 and at tf = 10s speed is
Vf = 2ms-1. As
acceleration=Slope of CD
=(change in velocity)/(change in time)=(v_f-v_i)/(t_f-t_i )=(2ms^(-1)-4ms^(-1))/(10s-5s)=(-2ms^(-1))/5s=-0.4ms^(-2)
Speed-time graph in figure 2.24 gives negative slope. Thus, the object has deceleration of 0.4 ms-2 .
DISTANCE TRAVELLED BY A MOVING OBJECT
The area under a speed-time graph represents the distance travelled by the object. If the motion is uniform then the area can be calculated using appropriate formula for geometrical shapes represented by the graph.
Example 2.9
A car moves in a straight line. The speed-time graph of its motion is shown in figure 2.25.
From the graph, find
(a) Its acceleration during the first 10 seconds.
(b) Its deceleration during the last 2 seconds.
(c) Total distance travelled.
(d) Average speed of the car during its journey.
Solution:
Acceleration during the first 10s is given as
a ⃑=(change in velocity)/(change in time)=(v_f-v_i)/(t_f-t_i )
=(16ms^(-1)-0)/(10s-0)=1.6ms^(-2)
Acceleration during the last two seconds is given as
a ⃑=(v_f-v_i)/(t_f-t_i )=(0-16ms^(-1))/(32s-30s)=-8ms^(-2)
Total distance travelled is given as
As we know area under the graph represents distance travelled so we can write
We have a trapezium (OABC)
So
Distance travelled is given as
=1/2 (base×hight)+(width×length)+1/2(base×hight)
=1/2 (10s×16ms^(-1) )+(16ms^(-1)×18s)+1/2(2s×16ms^(-1))
=(10s×8ms^(-1) )+(16ms^(-1)×18s)+(1s×16ms^(-1))
=(80m)+(288m)+(16m)
distance travelled=384m
Average speed is given as
Average Speed=(distance covered)/(total time taken)=384m/30s=12.8ms^(-1)
Required Answer.
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