When you normally drive the freeway between Sacramento and San Francisco at an average speed of 115 km/hr (71 mi/h ), the trip takes 1hr and 13 min . On a Friday afternoon, however, heavy traffic slows you down to an average of 90 km/hr (56 mi/h ) for the same distance. How much longer does the trip take on Friday than on the other days?

It takes 20 minutes and 24 seconds more on Friday than that of normal days.

Explanation:

Given:

Average speed on normal days (s₁) = 115 km/h = 71 mi/h

Time taken on normal days (t₁) = 1 hr and 13 min

Average speed on Friday (s₂) = 90 km/h = 56 mi/h

Time taken on Friday (t₂) = ?

Distance covered is same in both case.

Converting time from minutes to hours using the conversion factor, we have:

1 min = hour

∴ 13 min =

So,

Using the distance formula for both the days, we get:

Therefore,

So, time taken on Friday is 1.56 hours.

Time taken on Friday is more and the difference is given as:

Time difference = Time taken on Friday - Time on normal days

Time difference = 1.56 - 1.22 = 0.34 hours

Converting time from hours to minutes using the conversion factor, we have:

1 hour = 60 min

∴ 0.34 hours = 0.34 × 60

                      = 20.4

                      = 20 mins + 0.4 mins

                      = 20 min + (0.4 × 60) = 20 mins + 24 s

So, it takes 20 min and 24 seconds more on Friday than that of normal days.

at right angles to each other

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y