A student drives 4.8-km trip to school and averages a speed of 22.6 m/s. on the return trip home, the student travels with an average speed of 16.8 m/s over the same distance. what is the average speed (in m/s) of the student for the two-way trip? (be careful.)

In this case, we cannot simply take the average speed by adding the two speeds and divide by two.

What we have to do is to calculate the time required going to school and the return trip home.

We know that to calculate time, we use the formula:

t = d / v

where,

d = distance = 4.8 km = 4800 m

v = velocity

Let us say that the variables related to the trip going to school is associated with 1, and the return trip home is 2. So,

t1 = 4800 m / (22.6 m / s)

t1 = 212.39 s

t2 = 4800 / (16.8 m / s)

t2 = 285.71 s

total time, t = t1 + t2

t = 498.1 s

Therefore the total average velocity is:

= (4800 m + 4800 m) / 498.1 s

= 19.27 m / s = 19.3 m / s

Answer:

19.3 m/s