Q 11-02-118JEE MainJEE Main 2020 (5 Sep, Shift 2)Easy
The velocity ($v$) and time ($t$) graph of a body in straight line motion is shown in the figure. The point $S$ is at $4.333$ s. The total distance covered by the body in $6$ s is:
Answer: (A) $\dfrac{37}{3}$ m
Distance is the total area between the graph and the time axis (taking all areas as positive).
$0$ to $2$ s (triangle): $\tfrac12\times2\times4 = 4$ m
$2$ to $3$ s (rectangle): $1\times4 = 4$ m
$3$ to $4.333$ s (triangle): $\tfrac12\times\tfrac43\times4 = \tfrac83$ m
$4.333$ to $5$ s (triangle below axis): $\tfrac12\times\tfrac23\times2 = \tfrac23$ m
$5$ to $6$ s (triangle below axis): $\tfrac12\times1\times2 = 1$ m
$$s = 4 + 4 + \frac83 + \frac23 + 1 = \frac{37}{3}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics