Question

Question #22cd7

Answer 1

This is what I got

Explanation:

We are required to minimize time of journey. Hence, the car must

1. accelerate quickly from rest to maximum velocity, say in time `t_1`,
2. travel at maximum velocity some distance, say in time `t_2` and
3. retard from maximum velocity to rest in minimum possible time, say in time `t_3` to complete the journey.

`108\ kmhr^-1=108xx1000/3600=30\ ms^-1`

Applicable kinematic expressions are

`v=u+at` and `s=ut+1/2at^2`

1. `v_max=u+at_1`
   `=>t_1=(v_max-u)/a`
   `=>t_1=(30-0)/0.2=150\ s`
   In this time interval distance covered `s_1=1/2xx0.2xx(150)^2`
   `=>s_1=2250\ m`
2. We take up calculation of `t_3` first
   `t_3=(0-v_max)/r`
   `t_3=(0-20)/(-0.3)=66.bar6\ s`
   Distance covered in this time interval `s_3=20xx66.bar6+1/2xx(-0.3)xx(66.bar6)^2`
   `=>s_3=666.bar6\ m`
3. Remaining distance to complete the journey
   `s_2=6000-2250-666.bar6=3083.bar3\ m`
   Time taken to cover this distance at `v_max`
   `t_2=(3083.bar3)/30=102.bar7\ s`
   Total time `=t_1+t_2+t_3=319.bar4\ s`