# How do I solve this Kinematics question? [closed]

In 1987, Art Boileau won the Los Angeles Marathon, 26 miles and 285 yards, in 2h, 13 minutes and 9 seconds. At the 21-mile mark, Boileau had a 2.50 min lead on the second place winner, who later crossed the finish line 30 seconds after Boileau. Assume that Boileau maintained one constant average speed during the race and that both runners had been running at the same speed when Boileau passed the 21-mile marker. Find the average acceleration (in $m/s^2$) that the second place contestant had during the remaining part of the race after Boileau passed the 21-mile marker.

To solve this equation I first converted everything in meters and seconds

Total Race Length = 26 miles and 285 yards = 42195 meters

Time for Art to complete the race = 2 hours, 13 minutes and 9 seconds = 7989 seconds

21 miles = 33796 meters

I then identified Art's velocity = $\frac{42195m}{7989s}$ = 5.282 m/s

Next I identified after how long it takes for Art to reach the 21 mile mark:

$Time = \frac{Distance}{Velocity} \frac{33796 m}{5.282 m/s} = 6398.7701$ seconds

Let $d_x$ denote the position of the second place winner when Art hits the 21 mile mark. Let $a$ denote the second place winner's acceleration. Since it took, 150 seconds for the second place winner to travel the distance between $d_x$ and the 21 mile mark:

$33796 m - d_x = 5.282m/s(150s) + \frac{1}{2}a(150)^2$

which comes from the equation:

$d = v_it + \frac{1}{2}at^2$

Next I identified how long it took for Art to hit the 21 mile mark.

Time for Art to hit the 21 mile mark = $\frac{33796}{5.282m/s} = 6398.7701$ seconds

The time for Art to complete the remaining of the race from the 21 mile mark to the end = Time to complete the whole race - Time to hit the 21 mile mark = 7989 - 6398.7701 = 1590.2299 seconds.

It takes the second place winner 30 seconds more than 1590.2299 seconds to complete the race from his current location $d_x$. Therefore:

$42195 - d_x = 5.282m/s(1620.2299s) + \frac{1}{2}a(1620.2299)^2$

Now I have 2 linear equations:

$33796 m - d_x = 5.282m/s(150s) + \frac{1}{2}a(150)^2$

$42195m - d_x = 5.282m/s(1620.2299s) + \frac{1}{2}a(1620.2299)^2$

Solving them gives me $d_x = 32998.34275$ meters and $a = 4.87 * 10^{-4}m/s^2$. However the answer says the acceleration is $5.86 * 10^{-4} m/s^2$. What am I doing wrong?

-
 Hi Farhad - generally we discourage questions that just ask for someone to check your work. Once you have identified the specific concept that you're not sure about, that's the point at which it's appropriate to ask a question here. In this case, take a look at the work you've done and try to identify what concept you're not sure about, and make the question about that. Once you do, I'll be happy to reopen it. – David Zaslavsky♦ Sep 23 '12 at 0:15

## closed as too localized by David Zaslavsky♦Sep 23 '12 at 0:14

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