Answer :
Answer:
The distance of Jack's father's work from home is 22.5 miles.
Step-by-step explanation:
Given;
time it takes Jacks father at usual speed, t = 30 minutes = 0.5 hour
Jacks father speed when traffic is bad, = 30 mph slower
let the usual speed = v
let the distance of Jack's father's work from home = y
When traffic is normal;
[tex]v = \frac{y}{0.5 } ---- equation \ (1)[/tex]
When traffic is bad;
[tex]v - 30 \ mph = \frac{y}{1 \ + \ 0.5} \\\\v -30 = \frac{y}{1.5} \\\\v= 30 +\frac{y}{1.5} ----equation \ (2)[/tex]
Solve equation (1) and (2) together;
[tex]\frac{y}{0.5} = 30 \ + \ \frac{y}{1.5} \\\\y = 0.5(30 \ + \ \frac{y}{1.5})\\\\y = 15 + \frac{y}{3} \\\\y - \frac{y}{3} = 15\\\\\frac{3y-y}{3} =15\\\\\frac{2y}{3} =15\\\\2y = 3(15)\\\\2y = 45\\\\y = \frac{45}{2} \\\\y = 22.5 \ miles[/tex]
Therefore, the distance of Jack's father's work from home is 22.5 miles.
The required distance of Jack's father's work from home is y = 22.5miles.
Given that,
Jacks father drives to work in 30 minutes while driving at his usual speed.
When traffic is bad, he drives 30 miles per hour slower and the trip takes one hour longer.
We have to determine,
How far is Jack's father's work from home.
According to the question,
Let, the usual speed = v
And the distance of Jack's father's work from home = y
Jacks father at usual speed taking time, t = 30 minutes = 0.5 hour,
Jacks father speed when traffic is bad, = 30 mph slower.
The formula for speed is defined as = distance ÷ time.
Therefore,
Jacks father drives to work in 30 minutes while driving at his usual speed.
[tex]Speed = \dfrac{distance}{time}\\\\v = \dfrac{y}{0.5}[/tex]
Then,
When traffic is bad, he drives 30 miles per hour slower and the trip takes one hour longer.
[tex]Speed = \dfrac{distance}{time}\\\\v - 30mph = \dfrac{y}{1+0.5}\\\\v = \dfrac{y}{1+0.5} + 30[/tex]
Substitute the value of v from equation 1 to equation 2,
[tex]v = \dfrac{y}{1+0.5} + 30\\\\\dfrac{y}{0.5} = \dfrac{y}{1.5} + 30\\\\\dfrac{y}{0.5} = \dfrac{y +30(1.5)}{1.5}\\\\y (1.5) = 0.5(y +45)\\\\1.5y = 0.5y + 22.5 \\\\1.5y - 0.5y = 22.5\\\\1y = 22.5\\\\y = 22.5 \ miles[/tex]
Hence, The required distance of Jack's father's work from home is y = 22.5miles.
To know more about Distance click the link given below.
https://brainly.com/question/14036614