Question

Traveling math word problem - thank you! :)

Answer 1

We are given:

car and train leave at the same time

average velocity of car = 50 miles/hour

average velocity of train = 70 miles/hour

train arrives 2 hours early

Assuming variables and making equations!

let the time taken by the train = t hours

since the car arrived late, it took more time as compared to the train

time taken by the car = t + 2 hours

since the distance from Pasadena to Sacramento doesn't change, both the vehicles covered the same distance, d

Distance travelled by the car:

d = 50 $\frac{miles}{hour}$ * (t+2) hours                             [distance = velocity * time]

d = 50(t+2) miles

rewriting in terms of t

$t = \frac{d-100}{50}$

Distance travelled by the train:

d = 70 $\frac{miles}{hour}$ * t hours

d = 70t miles

rewriting in terms of t

$t = \frac{d}{70}$

now we have two expressions for t, both of which are equal because t is just the time taken by the train

Finding the distance:

$t = \frac{d-100}{50}$

$t = \frac{d}{70}$

because t is the same:

$\frac{d-100}{50} = \frac{d}{70}$

$\frac{d-100}{5} = \frac{d}{7}$

7(d - 100) = 5(d)

7d - 700 = 5d

(7d - 5d) - 700 = 0                                        [subtracting 5d from both sides]

2d = 700                                                       [adding 700 on both sides]

d = 350 miles                                               [dividing both sides by 2]      

The distance between Pasadena and Sacramento is 350 miles!