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(A) To cover a distance $S$ in minimum time,the motorcycle must accelerate at the maximum rate $\alpha = 5 \, m/s^2$ until it reaches a maximum veloci

Vedclass · Accepted Answer

$30$

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(A) To cover a distance $S$ in minimum time,the motorcycle must accelerate at the maximum rate $\alpha = 5 \, m/s^2$ until it reaches a maximum velocity $v$,and then immediately decelerate at the maximum rate $\beta = 10 \, m/s^2$ until it comes to rest.Let $t_1$ be the time of acceleration and $t_2$ be the time of retardation. The total time is $t = t_1 + t_2$.Using the kinematic equations:$v = \alpha t_1 = \beta t_2 \implies 5 t_1 = 10 t_2 \implies t_1 = 2 t_2$.The total distance $S = 1.5 \, km = 1500 \, m$ is given by the area under the velocity-time graph (a triangle):$S = \frac{1}{2} v (t_1 + t_2) = \frac{1}{2} (\alpha t_1) (t_1 + t_2) = \frac{1}{2} \left( \frac{\alpha \beta}{\alpha + \beta} ight) t^2$.Substituting the values:$1500 = \frac{1}{2} \left( \frac{5 	imes 10}{5 + 10} ight) t^2$$1500 = \frac{1}{2} \left( \frac{50}{15} ight) t^2 = \frac{1}{2} \left( \frac{10}{3} ight) t^2 = \frac{5}{3} t^2$.$t^2 = 1500 	imes \frac{3}{5} = 300 	imes 3 = 900$.$t = \sqrt{900} = 30 \, s$.

Column $I$	Column $II$
$(A)$ Constant positive acceleration	$(p)$ Speed may increase
$(B)$ Constant negative acceleration	$(q)$ Speed may decrease
$(C)$ Constant displacement	$(r)$ Speed is zero
$(D)$ Constant slope of $a-t$ graph	$(s)$ Speed must increase

The engine of a motorcycle can produce a maximum acceleration $5 \, m/s^2$. Its brakes can produce a maximum retardation $10 \, m/s^2$. What is the minimum time in which it can cover a distance of $1.5 \, km$?

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Match the following columns.

Column $I$ Column $II$

$(A)$ Constant positive acceleration $(p)$ Speed may increase

$(B)$ Constant negative acceleration $(q)$ Speed may decrease

$(C)$ Constant displacement $(r)$ Speed is zero

$(D)$ Constant slope of $a-t$ graph $(s)$ Speed must increase

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