From the v-t graph,find the total distance co | vedclass

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(A) The total distance covered is equal to the area under the $v-t$ graph.From the graph,the maximum velocity $V_{max}$ is reached at time $t_1$.We ha

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$\frac{1}{2} \left( \frac{\alpha \beta}{\alpha + \beta} \right) t^2$

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(A) The total distance covered is equal to the area under the $v-t$ graph.From the graph,the maximum velocity $V_{max}$ is reached at time $t_1$.We have $	an 	heta_1 = \alpha = \frac{V_{max}}{t_1} \implies t_1 = \frac{V_{max}}{\alpha}$.Similarly,$	an 	heta_2 = \beta = \frac{V_{max}}{t_2} \implies t_2 = \frac{V_{max}}{\beta}$.The total time is $t = t_1 + t_2 = V_{max} \left( \frac{1}{\alpha} + \frac{1}{\beta} ight) = V_{max} \left( \frac{\alpha + \beta}{\alpha \beta} ight)$.Thus,$V_{max} = \frac{\alpha \beta t}{\alpha + \beta}$.The total distance $S$ is the area of the triangle with base $t$ and height $V_{max}$:$S = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes t 	imes V_{max}$.Substituting $V_{max}$,we get $S = \frac{1}{2} 	imes t 	imes \left( \frac{\alpha \beta t}{\alpha + \beta} ight) = \frac{1}{2} \left( \frac{\alpha \beta}{\alpha + \beta} ight) t^2$.

From the $v-t$ graph,find the total distance covered by the car in time $t = t_1 + t_2$.

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