Question

Two trains travel at right angles to each other after leaving the same train station at the same time. Two hours later they are 4420 miles apart. If one travels 29 miles per hour slower than the other, what is the rate of the slower train?

Answer 1

Answer:1547.9 miles per hour

Step-by-step explanation:

Let the two trains leave in x direction and y direction

Velocity of train in x direction be v

therefore velocity of train in Y direction is v+29

because one train is 29 miles per hour slower than other.

Distance travel by train in x direction is(x) v(2)

Distance travel by train in y direction is(y) 2(v+29)

distance between them after 2 hours is 4420 miles

Now using pythagoras theorem

$x^2+y^2=\left ( 4420\right )^2$

$\left ( 2v\right )^2+\left ( 2\left ( v+29\right )\right )^2=\left ( 4420\right )$

on solving

$2v^2+58v-4881699=0$

v=$\frac{-58\pm \sqrt{\left ( 58 \right )^2+4\cdot 2\cdot4881699}}{2\cdot 2}$

v=1547.889 miles per hour