Car A has an initial speed of 15m/s and is gaining speed at 1.5m/s^2. Car B has an initial speed of 20m/s and is gaining speed at 2.0m/s^2. The initial distance between them is 168m. Find when and where the cars will pass one another and their speeds as they are passing.

Question

Distance for A
d = v1t + 1/2 a t^2
d = 0.75t^2 + 15t

Distance for B
d = v1t + 1/2 a t^2
d = t^2 + 20t

168-dB = dA
198 - t^2 + 20t = 0.75t^2 + 15t
168 - 1.75t^2 + 57 = 0
-1.75t^2 + 57 + 168 = 0

I don't know if that is right. I'm not sure how to find when so can someone please start me off on that?

Answer 1

ah, ha, found your errors while doing it. Two errors
"198 - t^2 + 20t = 0.75t^2 + 15t "
SHOULD BE
-168 + t^2 + 20 t = .75 t^2 + 15 t

well, if one is to catch up with the other, it must be car B catches car A because car B is going faster and is accelerating faster
therefore car B travels 168 meters more than car A
same t for both cars
Distance for A call it d
d = 15 t + (1.5/2) t^2
distance for B call it db
db = d+168 = 20 t + (2/2) t^2
or d = -168 + 20 t + t^2
so
-168 + 20 t + t^2 = 15 t + .75 t^2
.25 t^2 + 5 t - 168 = 0
t^2 + 20 t = 168*4
t ^2 + 20 t + 100 = 672 + 100
(t+10)^2 = 772
t = -10 +/-sqrt(772)
t = -10 +/- 27.8
t = 17.8 seconds because other root is <0
now find d and/or d+168

Answer 2

Sorry I didn't make this clear before. They're supposed to be passing each other instead of catching up. There's a diagram where both arrows are pointing to each other.

So d = 15 t + (1.5/2) t^2 is one equation. Would the other be a negative?

Answer 3

OH!!!
yes car a goes
d = 15 t + .75 t^2

car b goes 168 -d
168 -d = 20 t + t^2
so
d also = 168 - t^2 -20 t

so
168 - t^2 - 20 t = 15 t + .75 t^2
1.75 t^2 + 35 t -168 = 0

t = (1/3.5)[-35 +/- sqrt (1225 +4*1.75*168)]

t = (1/3.5)[ -35 +/- sqrt (2401)]
t = (1/3.5)[ -35 +/- 49]
use + 49 for positive time
t = 14/3.5 = 4 seconds