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 30^th to 35^th second, the motion retards and comes to a final part at the 35^th, 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 $a r e a = (\frac{1}{2}) \times s u m o f p a r a l l e l s i d e s \times a l t i t u d e .$
Substituting values from the graph,
$A r e a (D i s p l a c e m e n t) = \frac{1}{2} \times (P Q + O R) \times P T$
= $\frac{1}{2} \times [(30 - 10) + 35] \times 30$
= $\frac{1}{2} \times 55 \times 30$
= 825 m

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