Question

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.

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?

Answers

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
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?
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

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