Graphical Interpretation of Displacement, Velocity And Acceleration
Kinematics of Class 11
Average Velocity
The average velocity between two points in a given time interval can be obtained from a displacement versus time graph by computing the slope of the straight line joining the coordinates of the two points.
Instantaneous Velocity
The instantaneous velocity at time t is the slope of the tangent line drawn to the position−versus−time graph at that time.


Average Acceleration
The average acceleration between two points in a time interval is equal to the slope of the chord connecting the points on a velocity versus time graph.
Instantaneous Acceleration
The instantaneous acceleration at time t is the slope of the tangent drawn to the velocity versus time graph.


Displacement from Velocity Time GraphsGiven a velocity versus time graph, the displacement during an interval between time ti and tf is the area bounded by the velocity curve and the two vertical lines t = ti and t = tf, as shown in the Fig.(7.7 a). 



Velocity from Acceleration Time Graphs
Given an acceleration−versus−time graph, the change in velocity between t = ti and t = tf is the area bounded by the acceleration curve and the vertical lines t = ti and t = tf.


Table 7.1 Variation of Displacement (x), velocity (v) and acceleration (a) with respect to time for different types of motion.
Displacement 
Velocity 
Acceleration 

1. At rest 



2. Motion with constant velocity 



3. Motion with constant acceleration 



4. Motion with constant deceleration. 



Example: 7.4
At t = 0 a particle is at rest at the origin. Its acceleration is 2 m/s2 for the first 2 second and −2 m/s2 for the next 2 s. Plot the Solution
It is given that x = 0 and v = 0 at t = 0. The acceleration versus time graph is plotted in Fig. (7.9a). The velocity at t = 2 s is equal to the sum of velocity at t = 0 and the area under the acceleration−time graph between 

v_{2 }= v_{0} +
Since at t = 0: velocity is zero, vo = 0
∴ v_{2} = 0 + (2) (2) = 4 m/s
Similarly velocity at t = 4s is
v_{4 }= v_{0} + = 0 +2 (+2) + 2 (−2) = 0
Note that area below the x−axis is taken as negative
The displacement at t = 2 s is
x_{2 }= x_{0} +
∴ x_{2} = 0 + 1/2 (2) (4) = 4 m
and x_{4} = 0 + 1/2 (4) (4) = 8 m