A young woman named Kathy Kool buys a sports car that can accelerate at the rate of 4.16 m/s2. She decides to test the car by drag racing with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 0.58 s before Kathy. Stan moves with a constant acceleration of 2.88 m/s2 and Kathy maintains an accelera- tion of 4.16 m/s2.
Find the time it takes Kathy to overtake Stan. Answer in units of s
Find the distance she travels before she catches up with him. Answer in units of m
Find the speed of Stan’s car at the instant she overtakes him. Answer in units of m/s
Find Kathy’s speed at that instant. Answer in units of m/s
Let Kathy's distance from the starting point be x.
Let Stan's distance frm the starting point be y.
Let t = 0 be the instant that Stan starts. Kathy leaves 0.58 s later.
x = 2.08 (t-0.58)^2 (t > 0.58s)
y = 1.44 t^2
Set x = y and solve for t. Call that time t*
x(t*) is the distance travelled.
Stan's speed at that time is 2.88 t*
Kathy's speed at that time is 4.16(t*-0.58)
To solve this problem, we can use the equations of motion. Let's first find the time it takes for Kathy to overtake Stan.
1. Determine the time it takes for Stan to reach the point where Kathy starts her race:
We can use the equation of motion:
distance = initial velocity * time + 0.5 * acceleration * time^2
For Stan, the initial velocity is 0 m/s (starting from rest), the acceleration is 2.88 m/s^2, and we need to find the time it takes to cover this distance.
Let's denote this time as t1.
So we have:
0.58 s = 0 * t1 + 0.5 * 2.88 m/s^2 * t1^2
Simplifying the equation:
0.58 s = 1.44 m/s^2 * t1^2
Rearranging the equation:
t1^2 = 0.58 s / 1.44 m/s^2
t1^2 = 0.4028 s^2
Taking the square root on both sides:
t1 = √(0.4028 s^2)
t1 = 0.635 s (approx.)
Therefore, it takes Stan 0.635 seconds to reach the starting point of Kathy's race.
2. Determine the time it takes for Kathy to overtake Stan:
Now that Stan has used 0.635 seconds to reach Kathy's starting point, Kathy needs to "catch up" with him from her own starting point. Both cars start from rest, and Kathy has a higher acceleration.
Since Kathy starts 0.58 seconds after Stan, the time it takes for Kathy to overtake Stan is the difference between their times. Let's denote this time as t2.
t2 = t1 - 0.58 s
t2 = 0.635 s - 0.58 s
t2 = 0.055 s
Therefore, it takes Kathy 0.055 seconds to overtake Stan.
3. Find the distance she travels before she catches up with him:
To find the distance traveled by Kathy before overtaking Stan, we can use the equation of motion:
distance = initial velocity * time + 0.5 * acceleration * time^2
For Kathy, the initial velocity is 0 m/s (starting from rest), the acceleration is 4.16 m/s^2, and the time is 0.055 seconds.
Plugging in the values:
distance = 0 * 0.055 s + 0.5 * 4.16 m/s^2 * (0.055 s)^2
distance = 0.0006825 m
Therefore, Kathy travels approximately 0.0006825 meters before catching up with Stan.
4. Find the speed of Stan’s car at the instant she overtakes him:
We can find Stan's speed at the instant Kathy overtakes him by using the equation of motion:
final velocity = initial velocity + acceleration * time
For Stan, his initial velocity is 0 m/s (starting from rest), his acceleration is 2.88 m/s^2, and the time is 0.055 seconds (the same as Kathy's overtaking time).
Plugging in the values:
final velocity = 0 m/s + 2.88 m/s^2 * 0.055 s
final velocity = 0.1584 m/s
Therefore, at the instant she overtakes him, Stan's speed is approximately 0.1584 m/s.
5. Find Kathy’s speed at that instant:
To find Kathy's speed at the instant she overtakes Stan, we can use the equation of motion:
final velocity = initial velocity + acceleration * time
For Kathy, her initial velocity is 0 m/s (starting from rest), her acceleration is 4.16 m/s^2, and the time is 0.055 seconds (the same as Stan's overtaking time).
Plugging in the values:
final velocity = 0 m/s + 4.16 m/s^2 * 0.055 s
final velocity = 0.2288 m/s
Therefore, at the instant she overtakes Stan, Kathy's speed is approximately 0.2288 m/s.