quadratics/algebra question HEEEEEEELLLLLLLLLPPPPPPPPPP

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I need this as soon as possible!! PLEASE HELP!!

During a race, Runner A is travelling at a speed of 7.1m/s and is currently 12m in front of Runner B. Runner B is traveling at a speed of 6.0m/s and wants to catch up to Runner A. If Runner B starts to accelerate at a rate of 0.2m/s/s, how long in seconds will it take him to catch up to runner A? If there are 250 m remaining in the race,(relative to Runner A) who will win the race?

I have found 2 equations that I think will help figure this out.

7.1y+12=6y+1/2(0.2)y²
x=7.1y
 
Hello, caitlin!

During a race, Runner A is travelling at a speed of 7.1 m/s and is currently 12m in front of Runner B.
Runner B is traveling at a speed of 6.0 m/s and wants to catch up to Runner A.
If Runner B starts to accelerate at a rate of 0.2 m/s/s,
how long in seconds will it take him to catch up to runner A?
If there are 250 m remaining in the race,(relative to Runner A) who will win the race?
Click to expand...
Runner A has a constant speed of: \(\displaystyle \,V_{_A}\,=\,7.1\text{ m/s.}\)

Runner B is accelerating at 0.2 m/s<sup>2</sup>.
\(\displaystyle \;\;\)At time \(\displaystyle t\), his speed is: \(\displaystyle \,v_{_B}\:=\:6.0\,+\,0.2t\text{ m/s.}\)

At time \(\displaystyle t\), A's distance is: \(\displaystyle \,12\,+\,7.1t\) meters.
At time \(\displaystyle t\), B's distance is: \(\displaystyle \,(6.0\,+\,0.2t)t\) meters.

At time \(\displaystyle t\), B catches up to A, so their distances are equal:\(\displaystyle \;(6.0\,+\,0.2t)t\;=\;7.1t\)

We have a quadratic equation: \(\displaystyle \:0.2t^2\,-\,1.1t\,-\,12\:=\:0\)

Multiply by 10: \(\displaystyle \;2t^2\,-\,11t\,-\,120\:=\:0\)

Quadratic Formula: \(\displaystyle \:\,t\:=\:\frac{-(-11)\,\pm\,\sqrt{(-11)^2\,-\,(4)(2)(-120)}}{2(2)}\;=\;\frac{11\,\pm\,\sqrt{1081}}{4}\)

Therefore: \(\displaystyle \:t\;=\;10.96964111\:\approx\:11\) seconds.


In 11 seconds, A will run only: \(\displaystyle \,7.1(11)\:=\:78.1\) meters when B catches up and passes him.
\(\displaystyle \;\;\)B wins by a huge margin . . . especially if he keeps accelerating ... LOL!
 
Thanks

Hey thanks alot Soroban,

only thing is how did you get the quadratic equation from (6.0+0.2t)t=7.1t?

Also I made a graph with my data and according to the graph they intersetced at about 17.5s, do u think 11 s is realistic?
 
Small typo, but what follows is correct.
At time t, B catches up to A, so their distances are equal
(6+.2t)t = 7.1t + 12


Did you graph the two equations
(6+.2x)x and
7.1x+12?

PS. Soroban hyptonized me. Your equations are the right ones.
6t+.1t²=7.1t+12
t²-11t-120=0
t=(11+sqrt(11²+4*120))/2 =
16.5 seconds.
 
I made a chart first and added to the accelertion each time until they met then made a graph. I got that they would meet after 17.5 s and Runner B would be trvaeling at a speed of 9.5 m/s once he finaly passed runner A.

Only think is I was supposed to use equations not graphs. I don't really understand how Soroban got that quadratic quation and why she multiplied by 10.
 
He forgot (and I did too) that
ds = (6+.2t)dt so
s = 6t+(1/2)(.2t²)
The quadratic comes from the fact that the distances are equal when B catches A.
A is at 7.1t+12
(that is the same as your x=7.1t when you add the lead distance to both sides.)
B is at 6t+(1/2).2t² =
6t+.1t²
7.1t+12=6t+.1t²
.1t²-1.1t-12
Multiplying by 10 gets rid of the troublesome decimals.
t²-11t-120 = 0
 
help

hey thanks i tried the quadratic equation and got 15 s as y answer but my teacher said the answer is 17.75s could u please show me how to get that answer with that quadratic equation using the formula



thanks
 
for ax²+bx+c=0
x=(-b+sqrt(b²-4ac))/2a so
for t²-11t-120
t=(11+sqrt(11²+4*1*120))/2*1 =
(11+24.518)/2 =
35.5/2 =
17.75

(I thru out (-b-sqrt(b²-4ac))/2a 'cause negative time doesn't apply to the problem.)
 
o.k i understand it now except I don't get what sqrt is? and the format you wrote it in kinda confused me........sorry to keep bothering you but could u please explain it again (differently).


thanxs
 
actually i completely understand now thank you soooooooo mcuh for your help your a lifesaver.
 
Just FYI:

caitlin_needs_help said:
i'm sorry that i have a life and that i'm not the one sitting in front of my computer screen the whole day doing someone else's high school math problems....a "moronic loser" like you....I guess math is your pleasure? Have fun with that....that's how it was with everything else in your pathetic life right?
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