A young woman named Kathy Kool buys a sports car that can accelerate at the rate of 4.77 m/s2. She decides to test the car by dragging with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.04 s before Kathy. If Stan moves with a constant acceleration of 3.59 m/s2 and Kathy maintains an acceleration of 4.77 m/s2, find the time it takes Kathy to overtake Stan.
Calculate the distance she travels before she catches him.
Calculate the speed of Kathy's car at the instant she overtakes Stan.
Calculate the speed of Stan's car at the instant he is overtaken by Kathy.
To solve this problem, we can use the equations of motion to calculate the distance Kathy travels before she catches Stan, as well as the speeds of both cars at the moment of overtaking.
Step 1: Calculate the time it takes Kathy to overtake Stan.
Let's assume t is the time taken for Kathy to catch Stan.
Since Kathy starts 1.04 seconds after Stan, the time taken for Stan to reach the point Kathy starts from is (t + 1.04) seconds.
Using the equation of motion s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time, we can write two equations for Kathy and Stan.
For Kathy: sKathy = 0 + (1/2)(4.77)t^2
For Stan: sStan = 0 + (1/2)(3.59)(t + 1.04)^2
Step 2: Calculate the distance Kathy travels before catching Stan.
To find the distance, we need to equate the distances traveled by Kathy and Stan at the moment of overtaking.
Setting sKathy = sStan, we have:
(1/2)(4.77)t^2 = (1/2)(3.59)(t + 1.04)^2
Step 3: Solve the equation.
To solve this quadratic equation, let's expand and simplify:
4.77t^2 = 3.59(t^2 + 2.08t + 1.0816)
4.77t^2 = 3.59t^2 + 7.6812t + 3.904424
Combining like terms, we get:
0.18t^2 - 7.6812t - 3.904424 = 0
Step 4: Solve the quadratic equation using the quadratic formula.
The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
Using the values a = 0.18, b = -7.6812, and c = -3.904424, we can find the values of t.
Step 5: Calculate the distance Kathy travels before catching Stan.
Using the value of t calculated in the previous step, substitute it back into the equation of motion for Kathy:
sKathy = (1/2)(4.77)t^2
Step 6: Calculate the speed of Kathy's car at the moment she overtakes Stan.
To find Kathy's speed at the moment of overtaking, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
For Kathy: vKathy = 0 + (4.77)t
Step 7: Calculate the speed of Stan's car at the moment he is overtaken by Kathy.
For Stan: vStan = 0 + (3.59)(t + 1.04)
By following these steps, we can find the answers to the given questions.
To find the time it takes for Kathy to overtake Stan, we need to compare the distances they travel. The equation we can use is:
distance = initial velocity * time + 0.5 * acceleration * time^2
For Stan:
distance_s = 0 * t_s + 0.5 * 3.59 * t_s^2
For Kathy:
distance_k = 0 * t_k + 0.5 * 4.77 * t_k^2
Given that Kathy starts 1.04s after Stan, we can say that:
t_k = t_s - 1.04
Now, we need to find the time 't' when they meet, which means that the distances traveled by both must be equal:
distance_s = distance_k
0.5 * 3.59 * t_s^2 = 0.5 * 4.77 * (t_s - 1.04)^2
Simplifying the equation:
3.59 * t_s^2 = 4.77 * (t_s - 1.04)^2
Now, we can solve this equation to find the value of 't_s', which is the time it takes Kathy to overtake Stan.
To calculate the distance Kathy travels before she catches Stan, we can substitute the value of 't_s' into the equation for Kathy's distance:
distance_k = 0 * (t_s - 1.04) + 0.5 * 4.77 * (t_s - 1.04)^2
To find the speed of Kathy's car at the instant she overtakes Stan, we can differentiate the equation for Kathy's distance with respect to time and substitute the value of 't_s':
speed_k = acceleration_k * t_s
where acceleration_k = 4.77 m/s^2.
Similarly, to find the speed of Stan's car at the instant he is overtaken by Kathy, we can differentiate the equation for Stan's distance with respect to time and substitute the value of 't_s':
speed_s = acceleration_s * t_s
where acceleration_s = 3.59 m/s^2.
a. d1 = d2,
0.5a*t^2 = 0.5a(0.5t-1.04)^2,
0.5*3,59t^2 = 0.5*4.77(t-1.04)^2,
1.795t^2 = 2.385(t-1.04)^2,
0.7526t^2 = t^2-2.08t+1.082,
0.7526t^2 -t^2+2.08t =1.082,
-0,2474t^2 +2.08t - 1.082 = 0,
Use Quadratic formula and get:
t = 7.85 s.
b. d = 0.5*a*t^2,
d = 0.5*4.77*(7.85)^2 = 147 m.
c. V = 4.77*7.85 = 37.44 m/s.
d. V = 3.59 * (7.85+1.04) = 3i.9 m/s.