Question

The velocity (v) - time (1) graph of an object moving along a straight line is as shown in Fig. 2(b).29. Calculate the distance covered by object between (i) t = 0 to t = 5 s (ii) t = 0 to t = 10 s.​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Distance travelled in 0 to 5 s

= Area of trapezium OAB

= 1/2×(sum of parallel sides)×height

= 1/2×[(5-0)+(5-2)]×20 m

= 1/2×(5+3)×20 m

= 1/2×8×20 m

= 80 m

Likewise Distance travelled in 0 to 10 s

=Area of trapezium OAB + Area of triangle BC

=80 + 1/2×b×h

=80 + 1/2×(10-5)×20 m

=80 + 1/2×5×20 m

=80 + 50 m

=130 m