When you normally drive the freeway between Sacramento and San Francisco at an average speed of 115 km/h (71.5 mi/h), the trip takes 1 h and 13 min. On a Friday afternoon, however, heavy traffic slows you down to an average of 105 km/h (65.2 mi/h) for the same distance. How much longer does the trip take on Friday than on the other days?

∆t = 6.95 minutes

It takes 6.95 minutes longer on Friday than on the other days

Explanation:

Given:

Normal Average speed v1 = 115 km/h

Time taken t1 = 1 h 13 min = (1 + 13/60) hours = 1.2167 h

Distance travelled d = average speed × time taken

d = v1 × t1

Substituting the values;

d = 115 × 1.2167

d = 139.9205 km

On a Friday;

The average speed v2 = 105 km/h

Time taken = distance/average speed

t2 = d/v2

Substituting the values;

t2 = 139.9205/105

t2 = 1.332576190476

t2 = 1.3326 hours

The difference in time taken is;

∆t = t2 - t1

Substituting the values;

∆t = 1.3326 - 1.2167

∆t = 0.1159 hour

Converting to minutes;

∆t = 0.1159 × 60 minutes

∆t = 6.954 minutes

∆t = 6.95 minutes

It takes 6.95 minutes longer on Friday than on the other days

if you answer my question i will answer yours

answer; how far out of place which the object finished relative to the starting location;