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 [tex]\frac{miles}{hour}[/tex] * (t+2) hours [distance = velocity * time]
d = 50(t+2) miles
rewriting in terms of t
[tex]t = \frac{d-100}{50}[/tex]
Distance travelled by the train:
d = 70 [tex]\frac{miles}{hour}[/tex] * t hours
d = 70t miles
rewriting in terms of t
[tex]t = \frac{d}{70}[/tex]
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:
[tex]t = \frac{d-100}{50}[/tex]
[tex]t = \frac{d}{70}[/tex]
because t is the same:
[tex]\frac{d-100}{50} = \frac{d}{70}[/tex]
[tex]\frac{d-100}{5} = \frac{d}{7}[/tex]
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!
Answer:
350 miles
Step-by-step explanation:
Let :
Finding t
Substitute t in the equation formed for distance travelled by the train
The distance from the railroad station (Pasadena) to the state fair (Sacramento) is 350 miles
