Answer to Question #108894 in Mechanical Engineering for jyothi

Question

janes brother andrew leaves home for school at 8:00 am.he walks at 3.3kph.at 8:20 am jane discovers that andrew has left his homewirk at home .she decides to follow him on her bicycle and give him his homework,but she wants to be back home in time for an online test at 9:00 am .what is the minimum constant speed at which she needs to ride?

Answer 1

speed of andrew is 3.3 kph,

distance travel by andrew in 20 min = $speed\times time = 3.3 \times \frac {20}{60}= 1.1$ km

let at time t they will met , since speed of janes is constant so time take of going and coming will be same.

And total time to travel= 40 min so , time for going and coming , t = 20 min = $\frac{20}{60}= \frac {1}{3}$ hr

so, Let us take speed of janes be v

so,

distance cover by janes in 20 min = distance cover by andrew in 40 min

vt= $1.1+ 3.3 \times t$

$v \times \frac{1}{3}= 1.1 + 1.1 \times \frac{1}{3}$

v= $3\times 2.2 = 6.6 \frac {km}{h}$

this is the constant speed of janes