Question

A jet flies 425 km from Ottawa to Québec at rate v + 60. On the return flight, the

Answer 1

Answer:

a. $\frac{- 42,500}{(v + 60)(v - 40)}$

Step-by-step Explanation:

Given:

Distance Ottawa to Québec = 425 km

Initial flight rate = v + 60

Return flight rate = v - 40

$t = \frac{d}{r}$

Required:

Flight times difference of the initial and return flights

Solution:

=>Flight time of the initial flight:

$t = \frac{d}{r}$

$t = \frac{425}{v + 60}$

=>Flight time of the return flight:

$t = \frac{425}{v - 40}$

=>Difference in flight times:

$\frac{425}{v + 60} - \frac{425}{v - 40}$

$\frac{425(v - 40) -425(v + 60)}{(v + 60)(v - 40)}$

$\frac{425(v) - 425(40) -425(v) -425(+60)}{(v + 60)(v - 40)}$

$\frac{425v - 17000 -425v - 25500}{(v + 60)(v - 40)}$

$\frac{425v - 425v - 17000 - 25500}{(v + 60)(v - 40)}$

$\frac{- 42,500}{(v + 60)(v - 40)}$