Thursday

October 8, 2015
Posted by **Chris ** on Thursday, October 6, 2011 at 3:04pm.

- Math -
**Steve**, Thursday, October 6, 2011 at 3:46pmIf B wants to catch A in a mile, then that means he only has 1/100 hr.

Now, if we assume that B can continue to accelerate at the same rate he did to get to 60 mph, then that makes a =60mph/6sec = 10mph/sec = 10mi/(hr*sec) = 10mi(hr*1/3600hr) = 36000mi/hr²

At this point we are stuck. If B accelerates only to 100mph or less, he'll forever be behind A.

If he accelerates to some speed > 100mph, then there are many answers, depending on just how fast he ends up going.

If he accelerates at the same rate for the whole 36 seconds, then he ends up going v = at = 36000mi/hr² * 1/100 hr = 360 mi/hr

s = 18000 * 1/10000 = 1.8mi

So, if he accelerates constantly, he goes 1.8 mi in 1/100 hr. So, he catches A in less than 36 seconds.

So, how far does A get if a constantly accelerating B comes zipping up on his way to 360mph?

100t = 18000 t²

100 = 18000 t

t = 1/180 hr = 20 seconds.

So, the fastest A can catch B is 20 seconds, in 5/9 miles, having reached a speed of 200 mph.

Depending on acceleration, any time from

20 seconds on is possible.

- Math -
**Chris**, Thursday, October 6, 2011 at 4:11pmWould it be plausible for Car B to catch Car A within .5 mile?

- Math -
**Steve**, Friday, October 7, 2011 at 5:11amSo, in half the distance? No. Read above. Unless he can increase his acceleration, B cannot overtake A in less than 5/9 miles.