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Linear motion graphs with examples

Answer Text:
Motion graphs
Distance-time graphs

Area under velocity-time graph
Consider a body with uniform or constant acceleration for time ‘t’ seconds;

Distance travelled= average velocity x t = (0+v/2)xt =1/2vt
This is equivalent to the area under the graph. The area under velocity-time graph gives the distance covered by the body under‘t’ seconds.
Example
A car starts from rest and attains a velocity of 72km/h in 10 seconds. It travels at this velocity for 5 seconds and then decelerates to stop after
another 6 seconds. Draw a velocity-time graph for this motion. From the graph;
i. Calculate the total distance moved by the car
ii. Find the acceleration of the car at each stage.

Solution
a. From the graph, total distance covered= area of (A+B+C) = (1/2X10X20) + (1/2X6X20) + (5X20) =100+60+100 =260m
Also the area of the trapezium gives the same result.
b. Acceleration= gradient of the graph
c. Stage A gradient= #(20-0)/ (10-0)# = 2 #ms^-2#
d. Stage b gradient= #(20-20)/(15-10)# =0 #ms^-2#
e. Stage c gradient= #(0- 20)/(21-15)# =-3.33 #ms^-2#