Kathy has a sports car that can accelerate at the rate of 4.90m/s^2. She races with Stan. Both cars start from rest but experienced Stan leaves starting line 1.00s before her.
Stan moves with a constant aceleration of 3.90m/s^2 and Kathy maintains an acceleration of 4.90m/s^2
a) find time at which Kathy overtakes Stan
b) The distance she travels before she catches him
c) speed of both cars the instant she overtakes him
Are these the right equations??
Position as a function of time: xf= xi + vxi*t + 1/2*ax*t^2
velocity as a function of time: vxf= vxi + axt
3. The attempt at a solution
Well there was a similar example in the book but it involved a trooper and it started at vi=0 and chased a car with constant acceleration. The car was at a constant v and the trooper started chasing it after 1s after it passed a bilboard where the trooper was stationed. The thing is they gave the velocity and and since you knew that since it was 45.0m/s then after 1 s the initial distance of the car from the billboard was 45.0m and that was when the trooper started to move.
For this question I'm given acelleration for both. This has gotten me a bit confused as to how to go about finding the displacement of the cars and also since the initial v for both cars is 0.
Kathy
vi= 0m/s
a= 4.90m/s^2
Stan
vi= 0m/s
a= 3.50m/s^2
I don't know if it is correct but ...
-don't I need to find the position of Stan when Kathy starts to drive?
I don't know how to find that though.
For position as a function of time
xf= xi + vxi*t + 1/2 ax* t^2
xf Kathy= 0 + 0 + 1/2 (4.90m/s^2)(t^2)
xf Kathy= 1/2(4.90m/s^2)*(t^2)
Would it be the same for Stan? like this:
however stan starts 1.00 sec ahead of her...HOw do I visualize what I need for the problem?
I know I need the distance x which equates to both but how I include the 1.00 s is kind of confusing.
Is the time negative or positive? The book ex of the other one had a diagram like this:
t= -1.00s ------------t= 0s------------t= ?s
xf= xi + vxi*t + 1/2 ax* t^2
xf Stan= 0+ 0t + 1/2( 3.50m/s^2)(t^2)
Would I equate these 2 equations?? I'm trying to figure out how I incorperate the 1.00s since right now it is like they both started at the same time if I'm not incorrect.
Okay THANK YOU!!
To solve this problem, you can follow these steps:
1. Assign variables to the unknowns:
Let t be the time at which Kathy overtakes Stan.
Let xf_kathy be the position of Kathy at time t.
Let xf_stan be the position of Stan at time t.
2. Using the equations you provided, calculate the positions of Kathy and Stan at time t:
For Kathy:
xf_kathy = 1/2 * (4.90 m/s^2) * (t^2)
(Note that the initial position xi for both Kathy and Stan is assumed to be 0 since they start from rest.)
For Stan:
xf_stan = 1/2 * (3.90 m/s^2) * (t - 1.00 s)^2
(Note the t - 1.00 s in the equation to account for Stan starting 1.00 s before Kathy.)
3. Set up an equation for Kathy overtaking Stan:
Since Kathy overtakes Stan, their positions will be equal at time t.
xf_kathy = xf_stan
Substituting the expressions for xf_kathy and xf_stan, we get:
1/2 * (4.90 m/s^2) * (t^2) = 1/2 * (3.90 m/s^2) * (t - 1.00 s)^2
4. Solve the equation:
Simplify the equation and solve for t.
5. Once you have the value of t, you can calculate the distance Kathy travels before she catches Stan:
Use the equation for xf_kathy with the obtained value of t to find xf_kathy. Remember to subtract the initial position (xi = 0).
6. To find the speeds of both cars at the instant Kathy overtakes Stan:
Use the equation for velocity as a function of time: vxf = vxi + axt
For Kathy, vxf_kathy = 0 + (4.90 m/s^2) * t
For Stan, vxf_stan = 0 + (3.90 m/s^2) * (t - 1.00 s)
Evaluate these expressions using the value of t obtained in step 4.
Remember to double-check your calculations and units throughout the problem.
what you need here is to calibrate time.
Let t be the time Stan drives. Then t+1 is the time K drives. They both go the same distance, so equate them
K distance= Stan distnace
1/2 Ak (t+1)^2 = 1/2 As t^2
now solve for t.
First u have to find the distance between the place they overtake, and the starting point of the two racers, because the distances are the same. X(kathy)=2.46t^2.
X(stan)=1.75(t+1)^2 ... Then the only thing u can do is.
Xkathy =Xstan . And there u go the answer. T=5.46s ....