Distance, Rate, and Time Problem

Question

Because of bad weather, a bus driver reduced his usual speed along a 165 mile route by 5mph. The bus arrived 15 minutes later than usual. How fast does the bus usually travel?

Answer 1

Let x = usual speed of the bus

 

Then, x-5 = speed of bus in bad weather

 

Distance = (Rate)(Time), so Time = Distance/Rate

 

Time of bus at usual speed = Time of bus in bad weather  - 1/4

        (Note that 15 minutes = 1/4 hour)

 

So, 165/x = 165/(x-5) - 1/4

 

Multiply the equation by the LCD, 4x(x-5), to get:                                

 

     660(x-5) = 660x - x(x-5)

 

Simplifying, we obtain x² - 5x - 3300 = 0

 

By the quadratic formula, x = [5 ± √13225]/2 = 60 or -55

 

Since x can't be negative, x = 60 mph

 

 

Answer 2

This is solved by 2 equations and 2 unknowns:

 

 velocity *time = distance (remember time has to be expressed in hours because velocity is expressed as miles per hour.

 

Equation 1:

 

V*t=d or v*t=165 miles

 

Equation 2:

 

(v-5)(t+.25)=165 - this is basically stating velocity minus 5 mph multiplied by the time plus 15 minutes (.25 hours).

 

Now this is just 2 equations and 2 unknowns.  You can solve this by graphing and looking at the intersect or substition.

 

Using substition:

From first equation t = (165/v)

plugging into second equation:

 

(v-5)(165/v+.25)=165

FOIL and combine like terms (not shown)

825/v = .25*v  (cross multiply)

.25*v^2 = 825

v=57.4 mph (remember this is the original velocity).