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Question

Solve graphically. A body starting from rest is moving with an acceleration of 3 m/ s2 After 10 s, it continues to move with zero acceleration for 20 more seconds. It then retards and comes to a stop within the next 5 seconds. Find its total displacement. [3 MARKS]


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Solution

Graph : 1 Mark
Displacment : 2 Marks

We first draw the velocity-time graph.

Velocity at the end of 10 seconds when acceleration is 3 m/s2 is 30 m/s. P is the point representing this and OP is the portion of the velocity - time graph for this part of the journey, PQ represents the portion when the velocity is uniform and there is no acceleration. From the 30th to 35th second, the motion retards and comes to a final part at the 35th, second. QR represents this final part of the graph. At the point R, v is zero.

Thus OPQR represents the complete graph.

The total displacement is given by the area of OPQR.

Obviously it is trapezium,

Its area=(12)×sum of parallel sides×altitude.
Substituting values from the graph,
Area(Displacement)=12×(PQ+OR)×PT
= 12×[(3010)+35]×30
= 12×55×30
= 825 m


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