A motorbike starts from rest, attains a veloc | vedclass

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Question

(A) The motion of the motorbike is divided into three parts: acceleration, uniform velocity, and deceleration.$1$. Acceleration phase (from $A$ to $B$

Vedclass · Accepted Answer

$1070$

Vedclass · Answer

(A) The motion of the motorbike is divided into three parts: acceleration, uniform velocity, and deceleration.$1$. Acceleration phase (from $A$ to $B$):Initial velocity $u = 0$, final velocity $v = 10 \,m/s$, acceleration $a = 0.5 \,m/s^2$.Using $v = u + at$:$10 = 0 + 0.5 	imes t_{AB} \Rightarrow t_{AB} = \frac{10}{0.5} = 20 \,s$.$2$. Uniform velocity phase (from $B$ to $C$):Distance $d = 10 \,km = 10000 \,m$, velocity $v = 10 \,m/s$.Time $t_{BC} = \frac{d}{v} = \frac{10000 \,m}{10 \,m/s} = 1000 \,s$.$3$. Deceleration phase (from $C$ to $D$):Initial velocity $u = 10 \,m/s$, final velocity $v = 0$, deceleration $a' = 0.2 \,m/s^2$.Using $v = u - a't$:$0 = 10 - 0.2 	imes t_{CD} \Rightarrow 0.2 	imes t_{CD} = 10 \Rightarrow t_{CD} = \frac{10}{0.2} = 50 \,s$.Total time $T = t_{AB} + t_{BC} + t_{CD} = 20 \,s + 1000 \,s + 50 \,s = 1070 \,s$.

Column $I$	Column $II$
$(A)$ Change in velocity	$(p)$ $-5/3\,SI \text{ unit}$
$(B)$ Average acceleration	$(q)$ $-20\,SI \text{ unit}$
$(C)$ Total displacement	$(r)$ $-10\,SI \text{ unit}$
$(D)$ Acceleration at $t=3\,s$	$(s)$ $-5\,SI \text{ unit}$

$A$ motorbike starts from rest, attains a velocity of $10 \,m/s$ with an acceleration of $0.5 \,m/s^2$, travels $10 \,km$ with this uniform velocity, and then comes to a halt with a uniform deceleration of $0.2 \,m/s^2$. The total time of travel is: (in $\,s$)

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