The velocity (v) and time (t) graph of the body in straight line motion is shown in the figure. The point S is at 4.33 seconds. The total distance covered by the body in 6 second is:
Option: 1 \frac{11}{4}m
Option: 2 \frac{49}{4}m
Option: 3 \frac{37}{3}m
Option: 4 12m

Answers (1)

Total distance=area under the v-t graph=area of trapezium(OABS)+area of triangle=\frac{1}{2}(4.333+1) \times 4+\frac{1}{2} \times 1.667 \times 2= \frac{37}{3} m

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