Question

On Saturday morning, it took Jennifer 3.6 hours to drive to her mother's house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 hours to return home. Her speed was 5 mi/hr slower on Sunday than on Saturday. What was her speed on Sunday? Do not include units in your answer.

Answer 1

Answer:

• 45

Explanation:

We know that Jennifer's home and her mother's house are separated by a distance (in miles) d.

The average speed is

$v=\frac{distance}{time}$.

So, her Saturday average speed is:

$v_{saturday} = \frac{distance_{saturday}}{time_{saturday}}$

$v_{saturday} = \frac{d}{3.6 hours}$

And her Sunday average speed is

$v_{sunday} = \frac{distance_{sunday}}{time_{sunday}}$

$v_{sunday} = \frac{d}{4 hours}$

Now, we also know that

$v_{sunday} = v_{saturday} - 5 \frac{mi}{hr}$

Taking this in consideration, we can replace in the second equation

$v_{sunday} = \frac{d}{4 hours}$

$v_{saturday} - 5 \frac{mi}{hr} = \frac{d}{4 hours}$

$v_{saturday} = 5 \frac{mi}{hr} + \frac{d}{4 hours}$

Now, this is equal to the first equation:

$v_{saturday} = \frac{d}{3.6 hours} = 5 \frac{mi}{hr} + \frac{d}{4 hours}$

Working a little

$\frac{d}{3.6 hours} = 5 \frac{mi}{hr} + \frac{d}{4 hours}$

$\frac{d}{3.6 hours} - \frac{d}{4 hours} = 5 \frac{mi}{hr}$

$d (\frac{1}{3.6 hours} - \frac{1}{4 hours} )= 5 \frac{mi}{hr}$

$d * 0.02778 \frac{1}{hours} = 5 \frac{mi}{hr}  $

$d   =  \frac{5 \frac{mi}{hr}}{0.02778 \frac{1}{hours}}  $

$d   =  180 mi  $

Puting this in the second equation

$v_{sunday} = \frac{d}{4 hours}$

$v_{sunday} = \frac{180 mi}{4 hours}$

$v_{sunday} = 45 \frac{mi}{hours}$

Taking out the units

$v_{sunday} = 45$

Answer 2

The speed is basically the total distance covered in a given time interval.

Jennifer's speed while returning home on Sunday is 45 m/s.

How do you calculate the speed of Jennifer on Sunday?

Given that On Saturday morning, it took Jennifer 3.6 hours to drive to her mother's house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 hours to return home.

Let us consider that the distance between Jennifer's home and mother's home is d.

• Saturday

$s = \dfrac {d}{t}$

$s = \dfrac{d}{3.6}$.........equation 1

• Sunday

$s =\dfrac {d}{t}$

$s = \dfrac {d}{4}$.........equation 2

Her speed was 5 mi/hr slower on Sunday than on Saturday. This can be written as given below.

$\dfrac {d}{4} = \dfrac {d}{3.6} - 5$

By solving the above equation, we get the distance.

$-0.4d = -5(4\times 3.6)$

$d = 180 \;\rm m$

Thus, the speed on Sunday is calculated as given below.

$s = \dfrac {180}{4}$

$s = 45 \;\rm m/s$

Hence we can conclude that Jennifer's speed while returning home on Sunday is 45 m/s.

To know more about the speed and distance, follow the link given below.

https://brainly.com/question/12759408.