JEE PYQ: Motion in a Straight Line - Question ID 57f4a30c26ab (JEE Main 2020)

ID: 57f4a30c26abJEE Main 2020Single Correct MCQ
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 4.333 seconds. The total distance covered by the body in 6 s is : JEE Main 2020 (Online) 5th September Evening Slot Physics - Motion in a Straight Line Question 88 English
JEE Question illustration 57f4a30c26ab

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Step-by-step Explanation

4.333 sec = 133{{13} \over 3} sec

Distance = area under the v-t graph

= Area of Parallelogram + Area of triangle

= 12(4)(133+1){1 \over 2}\left( 4 \right)\left( {{{13} \over 3} + 1} \right) + 12(6133)×2{1 \over 2}\left( {6 - {{13} \over 3}} \right) \times 2

= 373{{37} \over 3} m

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