What average speed of the rest of the trip is needed if the car's overall average speed of the entire trip is 55.10 mph? Keep two digits after the decimal point.

Question

A car is making a trip in a total time of 98 min. During the first 34 min, its average speed is  24.5 mph.

Answer 1

First, let’s calculate the total distance traveled. This can be done using D = S x t, where S is the average speed in mph (55.10 mph or 0.918 miles per minute)  and t is the time in minutes. 

The total distance for the trip is 0.918 mpm x 98 min = 90.0 miles.



During the first part of the trip, the distance traveled is 24.5mph / 60 min x 34 min = 13.9 miles. This means that the remaining 76.1 miles must be traveled in 64 min. (98-34). The avg. speed to do this is 76.1miles / 64 min x 60 min/hr.= 71.3 mph.

Please check math.

Answer 2

Distance for 1st part of trip s₁ = 24.5mi/h ·34/60 h = 13.88 mi

Time fo rest of trip: t₂= 98 min - 34 min = 64 min = 1.07 h. Total distance s = 55.10 mi/h·98/60h = 90.00 mi;

Rest of trip distance: 90.00 - 13.88 mi = 76.12 mi. Average speed on rest of trip v_av2 = 76.12/1.07 = 71.14mph