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(A) For interval $(a)$ $t = 0\; s$ to $10\; s$:Distance = Area under the speed-time graph = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \fr

Vedclass · Accepted Answer

$6\; m/s$ and $9\; m/s$

Vedclass · Answer

(A) For interval $(a)$ $t = 0\; s$ to $10\; s$:Distance = Area under the speed-time graph = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 10\; s 	imes 12\; m/s = 60\; m$.Average speed = $\frac{	ext{Total distance}}{	ext{Total time}} = \frac{60\; m}{10\; s} = 6\; m/s$.For interval $(b)$ $t = 2\; s$ to $6\; s$:The speed $v(t)$ is given by $v(t) = 2.4t$ for $0 \le t \le 5$ and $v(t) = 12 - 2.4(t-5)$ for $5 At $t = 2\; s$,$v(2) = 2.4 	imes 2 = 4.8\; m/s$.At $t = 5\; s$,$v(5) = 12\; m/s$.At $t = 6\; s$,$v(6) = 12 - 2.4(6-5) = 9.6\; m/s$.Distance from $t = 2\; s$ to $5\; s$ = Area of trapezoid = $\frac{1}{2} 	imes (4.8 + 12) 	imes (5 - 2) = \frac{1}{2} 	imes 16.8 	imes 3 = 25.2\; m$.Distance from $t = 5\; s$ to $6\; s$ = Area of trapezoid = $\frac{1}{2} 	imes (12 + 9.6) 	imes (6 - 5) = \frac{1}{2} 	imes 21.6 	imes 1 = 10.8\; m$.Total distance = $25.2 + 10.8 = 36\; m$.Average speed = $\frac{36\; m}{6\; s - 2\; s} = \frac{36}{4} = 9\; m/s$.

The speed-time graph of a particle moving along a fixed direction is shown in the figure. What is the average speed of the particle over the intervals: $(a)$ $t = 0\; s$ to $10\; s$,and $(b)$ $t = 2\; s$ to $6\; s$?

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