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JEE Main 20205 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Evening Shift
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2020 · 5 September · Evening Shift · Physics · Motion in One Dimension

The velocity ( v ) and time ( t ) graph of a body in a straight...

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 :

The velocity ( v ) and time ( t ) graph of a body in a straight line motion is shown in the figure. The point S is at 4

Options

  1. A373m\dfrac{37}{3}m
  2. B12m12m
  3. C11m11m
  4. D494m\dfrac{49}{4}m

Correct answer

A. 373m\dfrac{37}{3}m

Step-by-step solution

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)⇒ 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⇒ d = 4 + 4 + \dfrac{8}{3} + \dfrac{5}{3}

⇒d=373m⇒ d = \dfrac{37}{3}m

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