rate problem

[ddadams]

Bobby and Rick are in a 10-lap race on a one mile oval track. Bobby averaging 90mph, has completed two laps just as Rick is getting his car onto the track. What speed does Rick have to average to be even with bobby at the end of the tenth lap?
*Hint: Bobby does 8 miles in the same time as Rick does 10 miles.

I am trying to find the equation:

rxt=d
rate of rick
rate of bobby= 90
distance =distance: 10=10

time of bobby
(b)90=10
90b=10
b=9
rick
I really can't think of the equation

 

[mmm4444bot]

… Hint: Bobby does 8 miles in the same time as Rick does 10 miles …

Hello DD:

That's a nice hint.

It makes a statement about the amount of time that it takes Bobby to drive 8 miles (while averaging 90 mph).

If we knew this amount of time, then we could use it to find Rick's rate. So, how much time is it?

(Note the asterisk used as a multiplication sign below. :idea: In algebra, the letter x represents a variable, so we no longer use it as a multiplication sign; scientific notation is one exception.)

d = r * t

8 = 90 * t

Solve this equation for t, and you'll have the fractional part of one hour that it takes Bobby to drive 8 miles.

Now, what does the hint say about this amount of time? It says that Rick drives 10 miles during the same amount of time.

d = r * t

10 = r * (substitute your result for t here)

Solve this equation for r, and that's the rate for which this exercise asks.

Cheers ~ Mark

 

[soroban]

Hello, ddadams!

Bobby and Rick are in a 10-lap race on a one mile oval track.
Bobby averaging 90 mph, has completed two laps just as Rick is getting his car onto the track.
What speed does Rick have to average to be even with bobby at the end of the tenth lap?
. . * Hint: Bobby does 8 miles in the same time as Rick does 10 miles.

\(\displaystyle \text{We know that: }\;\text{Distance} \:=\:\text{speed} \times \text{time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}\)

\(\displaystyle \text{Bobby drives 8 miles at 90 mph. }\;\; \text{ This takes: }\:\frac{8}{90}\text{ hours.}\)

\(\displaystyle \text{Rick drives 10 miles at }x\text{ mph. }\;\; \text{This takes: }\:\frac{10}{x}\text{ hours.}\)

. . \(\displaystyle \text{The times are equal: }\;\frac{8}{90} \:=\:\frac{10}{x} \quad\Rightarrow\quad 8x \:=\:900 \quad\Rightarrow\quad x \:=\:112.5\text{ mph}\)

 

[ddadams]

Thanks! I got it. so instead of the whole lenght of the track I can take the 8 mile hint and us that to find the times.

I appreciate it!