2 people are racing. What is the minimum length that person 1 needs to race in order to win.?

Question

person a: accelerates at 8m/s^2, and has a peak velocity of 12m/s.
person b: accelerates at 5m/s^2, and has a peak veloctiy of 16m/s.

What is the min. length of race that person b should agree to in order to win?

Does anybody know the equation for figuring this out?

by length i mean the distance in meters. thx for the help so far!

Answer 1

Clue
make them equal to each other to find out when they would exactly tie

Answer 2

The first part of this problem is to get a and b up to max velocity

then calculate the point when they will tie, and add a second to the time.

runner a will reach peak velocity in 1.5s2 (12 / 8)
Distance travelled by runner a in 1.5seconds = 8m + .5 x 12 m = 6) = 15m

Time taken for runner b to reach peak=16/5=3.2seconds
runner b will travel 5 m in first second, then 10 meters per second in the next second, plus the first 5 m = 15m - thus by second two, he is half a second behind runner a

meanwhile, runner a at 2 seconds has now run 15m accelerating, and half a secon of full speed of 12m/s / 2 = 6 m thus he has rum 15 + 6 = 21 m

by second three, runner b has run the 5m of second one, the 10m of second two and the 15 m of second three = 30m

runner a is at full speed, so he has added another 12 meters to his distance of 21m = 33m

By second 4, Runner b accelerates for 0.2 seconds, = ((15/5 = 2m) + (5m/s / 5 =1 m) = 3m) = 0.8 seconds at full speed = 16/10 x 8 = 12.8m

thus runner b has traveled 15.8 meters, and is 48.8 m from the start.

So 4 seconds should do it.

Answer 3

Time taken for person b to reach peak=16/5=3.2s
Time for person a to reach peak=12/8=1.5s
Distance travelled by a in1.5s=0.5x8x1.5²
=9m
Distance travelled by a in 3.2s=9+1.7x12
=29.4
Distance by b during 3.2s=0.5x5x3.2²
=25.6m
Difference=29.4-25.6
=3.8m
For b to win,ASSUMING he must at least win by 0.1m,hence he need to catch up to 3.9m

Let t be the time when b is winning(3.2s is excluded,
16t-12t=3.9
4t=3.9
t=0.975s

Hence B should agree to challenge until (3.2+0.975)
=4.175s

The distance=25.6+0.975x16
=41.2m (They meet at 40.8m)