question_answer

The engine of a motor cycle can produce maximum acceleration \[5\text{ }m/se{{c}^{2}}\]. Its breaks can produce a maximum retardation \[10\text{ }m/se{{c}^{2}}\]. What is minimum time in which it can cover a distance of 1.5 km:

A) 5 sec.

B) 10 sec.           

C) 15 sec.

D) 30 sec.

Correct Answer: D

Solution :

[d] For \[{{t}_{\min }}\] motion of particle is shown in \[v-t\] graph is
For accelerating motion \[\frac{{{v}_{0}}}{{{t}_{0}}}=5\]	...(i)
For retarding motion \[\frac{{{v}_{0}}}{T-{{t}_{0}}}=10\]	...(ii)
Distance = area of \[v-t\] curve
\[\frac{1}{2}{{v}_{0}}T=1500\]	...(iii)
From equation (i), (ii) & (iii) \[T=30\sec \] Ans.