MATH SOLVE

4 months ago

Q:
# Maria drove from chicago, illinois, to milwaukee, wisconsin, a distance of 90 miles, at a mean speed of 56 miles per hour. on her return trip, the traffic was much heavier, and her mean speed was 49 miles per hour. find maria's mean speed for the round trip. hint: divide the total distance by the total time. (round the answer to one decimal place.)

Accepted Solution

A:

- first trip: the distance is

S=90 m

while the speed is

v=56 m/h

Therefore the time taken is

[tex]t_1= \frac{S}{v_1}= \frac{90 m}{56 m/h}= 1.6 h[/tex]

- second trip: the distance covered is still the same,

S=90 m

while the average speed this time was

v=49 m/h

Therefore, the time taken for the second trip was

[tex]t_2 = \frac{S}{v_2}= \frac{90 m}{49 m/h}=1.8 h [/tex]

- The average speed for the whole trip is the total distance divided by the total time taken:

[tex]v= \frac{S_1 + S_2 }{t_1+t_2}= \frac{90 m+90 m}{1.6 h+1.8 h}=52.9 m/h [/tex]

S=90 m

while the speed is

v=56 m/h

Therefore the time taken is

[tex]t_1= \frac{S}{v_1}= \frac{90 m}{56 m/h}= 1.6 h[/tex]

- second trip: the distance covered is still the same,

S=90 m

while the average speed this time was

v=49 m/h

Therefore, the time taken for the second trip was

[tex]t_2 = \frac{S}{v_2}= \frac{90 m}{49 m/h}=1.8 h [/tex]

- The average speed for the whole trip is the total distance divided by the total time taken:

[tex]v= \frac{S_1 + S_2 }{t_1+t_2}= \frac{90 m+90 m}{1.6 h+1.8 h}=52.9 m/h [/tex]