Question

2. A student drives 7.8-km trip to school and averages a speed of

32.6 m/s. On the return trip home, the student travels with an
average speed of 36.8 m/s over the same distance. What is the
average speed (in m/s) of the student for the two-way trip? (Be
careful.) Also, what is the total time taken for the trip?

Answer 1

The total time is: t=451.22 sec

The average speed is: V=34.57 m/s

Step-by-step explanation:

Average speed

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

`\displaystyle V=(x)/(t)`

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

`\displaystyle t1=(x1)/(v1)=(7,800)/(32.6)=239.26\ sec`

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

`\displaystyle t2=(x2)/(v2)=(7,800)/(36.8)=211.96\ sec`

The total time is:

`t=239.26\ sec+211.96\ sec=451.22\ sec`

`\boxed{t=451.22\ sec}`

The average speed is:

`\displaystyle V=(15,600)/(451.22)=34.57\ m/s`

`\boxed{\displaystyle V=34.57\ m/s}`