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(B) The distance traversed is equal to the area under the speed-time graph.For the interval $t = 2\,s$ to $t = 5\,s$,the speed $v$ at time $t$ is give

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) The distance traversed is equal to the area under the speed-time graph.For the interval $t = 2\,s$ to $t = 5\,s$,the speed $v$ at time $t$ is given by $v = \frac{12}{5}t = 2.4t$.At $t = 2\,s$,$v_1 = 2.4 	imes 2 = 4.8\,m/s$.At $t = 5\,s$,$v_2 = 12\,m/s$.The area of the trapezoid formed between $t=2\,s$ and $t=5\,s$ is $S_1 = \frac{1}{2} 	imes (v_1 + v_2) 	imes \Delta t = \frac{1}{2} 	imes (4.8 + 12) 	imes (5 - 2) = \frac{1}{2} 	imes 16.8 	imes 3 = 25.2\,m$.For the interval $t = 5\,s$ to $t = 6\,s$,the speed $v$ at time $t$ is given by $v = 12 - \frac{12}{5}(t - 5) = 24 - 2.4t$.At $t = 5\,s$,$v_2 = 12\,m/s$.At $t = 6\,s$,$v_3 = 12 - 2.4(6 - 5) = 9.6\,m/s$.The area of the trapezoid formed between $t=5\,s$ and $t=6\,s$ is $S_2 = \frac{1}{2} 	imes (v_2 + v_3) 	imes \Delta t = \frac{1}{2} 	imes (12 + 9.6) 	imes (6 - 5) = \frac{1}{2} 	imes 21.6 	imes 1 = 10.8\,m$.Total distance $S = S_1 + S_2 = 25.2 + 10.8 = 36\,m$.

The speed-time graph of a particle moving along a fixed direction is shown in the figure. The distance traversed by the particle between $t = 2\,s$ to $t = 6\,s$ is........$m$

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