question_answer

A motor bike accelerates from rest at a constant rate of \[3\text{ }m{{s}^{-2}}\] for some time and then moves with uniform velocity for the same duration. Then it retards at a constant rate of \[6\text{ }m{{s}^{-2}}\] and comes to rest. The bike was in motion for a period of 5 seconds. The total distance travelled by the bike is:

A) 21 m

B) 24 m

C) 27 m    

D) 30 m

Correct Answer: A

Solution :

a) In \[v=u+at\] or \[v=u-at,\] if \[\left| \,a\, \right|\] is doubled, the time t is halved.

Hence, if t is the time for acceleration, then \[\frac{t}{2}\] is the time taken for retarded motion/

\[\therefore \] total time duration for travels is\[t+t+\frac{t}{2}=\frac{5t}{2}\]

\[\frac{5t}{2}=5s(given)\]

\[\therefore t=2s\]

(i) Distance covered in accelerated motion

\[{{S}_{1}}=ut+\frac{1}{2}a{{t}^{2}}\]

\[=0\,\,\times \,\,t+\frac{1}{2}\,\,\times \,\,3\,\,\times \,\,{{2}^{2}}=6m\]

(ii) Distance covered in uniform motion

\[v=u+at=0+3\times 2=6\,m{{s}^{-1}}\]

\[\therefore {{S}_{2}}=vt=6\times 2=12\,m\]

(iii) Distance travelled in retarded motion

\[{{S}_{3}}=ut-\frac{1}{2}\,\,a{{t}^{2}}\]

\[=6\times 1-\frac{1}{2}\,\,\times \,\,6\,\,\times \,\,{{1}^{2}}=3m\]

\[\therefore \]total distance traversed

\[S={{S}_{1}}+{{S}_{2}}+{{S}_{3}}=6+12+3=21\,\,m\]