Question

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?

Answer 1

Answer:

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 = $\frac{1}{60}$ hour

∴ 13 min = $\frac{13}{60}=0.22\ h$

So, $t_1=1\ h+0.22\ h=1.22\ h$

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

$d=speed\times time\\\\d=s_1t_1\\d=s_2t_2$

Therefore,

$s_1t_1=s_2t_2\\\\115\ km/h\times 1.22\ h=90\ km/h\times t_2\\\\t_2=\frac{140.3\ km}{90\ km/h}\\\\t_2=1.56\ hrs$

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.