Question

A tour bus normally leaves for its destination at 5:00p.m. for a 378 mile trip. This week however, the bus

leaves at 6:00 p.m. To arrive on time, the driver drives 9 miles per hour faster than normal What is the normal
speed of the bus?​

Answer 1

Explanation:

we know D = RT, where

D is distance

R is rate/speed

T is time

Normally:

D = RT

378 = RT

This week:

D = RT

378 = (R+9)(T-1)

These are the 2 equations we can write:

Let's mutliply out the last equation:

378 = (R+9)(T-1)

378 = RT - R + 9T - 9

387 = RT - R + 9T

we know RT = 378 and T = 378/R, we replace:

387 = 378 - R + 9(378/R)

9 = -R + 3402/R

Now solving:

R can't be negative, so we take R = 54 miles per hour

Answer 2

54 mph

Explanation:

we know D = RT, where

D is distance

R is rate/speed

T is time

Normally:

D = RT

378 = RT

This week:

D = RT

378 = (R+9)(T-1)

These are the 2 equations we can write:

Let's mutliply out the last equation:

378 = (R+9)(T-1)

378 = RT - R + 9T - 9

387 = RT - R + 9T

we know RT = 378 and T = 378/R, we replace:

387 = 378 - R + 9(378/R)

9 = -R + 3402/R

Now solving:

`9 = -R + (3402)/(R)\\R-(3402)/(R)+9=0\\R^2+9R-3402=0\\R=-63,54`

R can't be negative, so we take R = 54 miles per hour