Kinematics

JEE Main level, single correct, medium

101 of 308

Question 101 of 308

The velocity (v)(v) and time (t)(t) graph of a body in a straight line motion is shown in the figure. The point SS is at 4.3334.333 seconds. The total distance covered by the body in 6s6s is :

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Answer: (A) 373m\dfrac{37}{3}m

Total distance covered by the body is equal to the modulus of area under graph velocity-time graph
⇒d=(12×4×2)+(4×1)+(12×4×43)+(12×2×53)\Rightarrow d=\left(\dfrac{1}{2}\times 4\times 2\right)+\left(4\times 1\right)+\left(\dfrac{1}{2}\times 4\times \dfrac{4}{3}\right)+\left(\dfrac{1}{2}\times 2\times \dfrac{5}{3}\right)
⇒d=4+4+83+53\Rightarrow d=4+4+\dfrac{8}{3}+\dfrac{5}{3}
⇒d=373m\Rightarrow d=\dfrac{37}{3}m

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