Answer to Question #119566 in Physics for Joseph Wisdom

Question

A man traveled a distance of 120km by car from hisme to Lagos. On his way back, his average speed was increased by 5km/h and he found out that he took 20mins less
a. What was his average speed for the first part of the journey?
b. How long did he take for the double journey

Answer 1

Let the speed of a car for the first part of the journey is $v_1$. Then time of motion for the first part of the journey

$t_1=\frac{120}{v_1}$

For the second part we get

$t_1-\frac{1{\rm hr}}{3}=\frac{120}{v_1+5}$

Hence

$\frac{120}{v_1}-\frac{1}{3}=\frac{120}{v_1+5}$

$\frac{120}{v_1}-\frac{120}{v_1+5}=\frac{1}{3}$

Root

$v_1=40\:\rm km/hr$

Thus

$t_1=\frac{120}{40}=3\:\rm hr$

$t_2=t_1-1/3=3-1/3=2\rm\: hr\:40\: min$

Total time of motion

$t=t_1+t_2=5\rm\: hr\:40\: min$

Answers:

(a) $40\:\rm km/hr$

(b) $5\rm\: hr\:40\: min$