The figure shows a velocity-time graph of a p | vedclass

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(A) The total distance travelled is the sum of the magnitudes of the areas under the velocity-time graph with respect to the time axis.The velocity be

Vedclass · Accepted Answer

$66.7$

Vedclass · Answer

(A) The total distance travelled is the sum of the magnitudes of the areas under the velocity-time graph with respect to the time axis.The velocity becomes zero at $t = 4 + \frac{10}{15} = \frac{14}{3}$ s $\approx 4.67$ s.Area $1$ (from $t=0$ to $t=14/3$ s): The region is a trapezium with parallel sides $(4-2) = 2$ s and $14/3$ s, and height $10$ m/s. Area $= \frac{1}{2} 	imes \left(2 + \frac{14}{3}ight) 	imes 10 = \frac{100}{3} \approx 33.33$ m.Area $2$ (from $t=14/3$ to $t=8$ s): The triangle has base $\left(8 - \frac{14}{3}ight) = \frac{10}{3}$ s and height $|-20| = 20$ m/s. Area $= \frac{1}{2} 	imes \frac{10}{3} 	imes 20 = \frac{100}{3} \approx 33.33$ m.Total distance $= 33.33 + 33.33 = 66.67$ m.

The figure shows a velocity-time graph of a particle moving along a straight line. The total distance travelled by the particle is ........ $m$.

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