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?
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
To find when the two cars will pass each other, we need to solve the quadratic equation derived from the distances of the cars.
The equation we have is:
-1.75t^2 + 57 + 168 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -1.75, b = 57, and c = 168.
Plugging in these values into the formula, we get:
t = (-57 ± √(57^2 - 4(-1.75)(168))) / (2(-1.75))
Simplifying further:
t = (-57 ± √(3249 + 1176)) / -3.5
t = (-57 ± √(4425)) / -3.5
Now, we need to calculate the value inside the square root:
√(4425) ≈ 66.54
We have two possible solutions:
t1 = (-57 + 66.54) / -3.5
t2 = (-57 - 66.54) / -3.5
Calculating these values:
t1 ≈ 0.85 seconds
t2 ≈ -34.17 seconds
Since time cannot be negative in this context, we discard the negative solution.
Therefore, the cars will pass each other approximately 0.85 seconds after they start.
Next, we can find the distances traveled by each car at this time:
Distance traveled by Car A:
dA = 0.75t^2 + 15t
dA ≈ 0.75(0.85)^2 + 15(0.85)
dA ≈ 0.5419 + 12.75
dA ≈ 13.29 meters
Distance traveled by Car B:
dB = t^2 + 20t
dB ≈ (0.85)^2 + 20(0.85)
dB ≈ 0.7225 + 17
dB ≈ 17.72 meters
Therefore, the cars will pass each other approximately 0.85 seconds after they start, with Car A traveling about 13.29 meters and Car B traveling about 17.72 meters.