Question

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.)

Answer 1

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

Answer 2

Answer: 19.3 m/s

Step-by-step explanation:

Given : Distance between home and school = 4.8 km= 4800 meters

(∵ 1 km= 1000 meters)

Formula to calculate time :

$\text{Time} =\dfrac{\text{Distance}}{\text{Speed}}$

From home to school ,

Distance= 4800 m

Speed = 22.6 m/s.

$\text{Time} =\dfrac{4800}{22.6}=\dfrac{24000}{113}\ sec$

From school to home ,

Distance= 4800 m

Speed = 16.8 m/s.

$\text{Time} =\dfrac{4800}{16.8}=\dfrac{12000}{41}\ sec$

Now, the average speed (in m/s) of the student for the two-way trip is given by :-

$\dfrac{\text{Total distance}}{\text{Total time}}\\\\=\dfrac{4800+4800}{\dfrac{24000}{113}+\dfrac{12000}{14}}\\\\=\dfrac{9600}{498.10}\approx19.3\ m/s$

Hence, the average speed (in m/s) of the student for the two-way trip= 19.3 m/s .