A speeder doing 50.0 mi/hr (about 22 m/s) in a 25 mi/hr zone approaches a parked police car. The instant the speeder passes the police car, the police begin their pursuit. If the speeder maintains a constant velocity, and the police car accelerates with a constant acceleration of 4.50 m/s^2, (a) how long does it take for the police car to catch the speeder, (b) how far have the two cars traveled in this time, and (c) what is the velocity of the police car when it catches the speeder?
Now this is very similar but easier than the one you threw up from the roof of the building. You try it. I have grand kids playing hockey now.
I figured it out. i juss had to find the T to help me solve the other parts
Vs = 50mi/h * 1600m/mi * 1h/3600s = 22.22 m/s. = Velocity of the speedster.
a. Ds = Dp
22.2*t = 0.5a*t^2
Divide both sides by t:
0.5a*t = 22.2
0.5*0.450t = 22.2
0.225t = 22.2
t = 98.8 s.
b. Ds = Dp = 22.2*t = 22.2 * 98.8 = 2193 m.
c. V^2 = Vo^2 + 2a*d
Vo = 0
a = 0.450 m/s^2
d = 2193 m.
Solve for V.
To solve this problem, we can use the equations of motion. Let's consider the motion of the police car relative to the speeder.
(a) To find the time it takes for the police car to catch the speeder, we need to determine the time at which both cars are at the same position. We can do this by setting up the following equation:
Distance of the police car = Distance of the speeder
The distance for the police car is given by:
Distance = Initial velocity * time + (1/2) * acceleration * time^2
Since the police car starts from rest, the initial velocity is zero. Therefore, the equation for the distance of the police car simplifies to:
Distance of the police car = (1/2) * acceleration * time^2
The distance for the speeder is simply the product of their velocity and time:
Distance of the speeder = Speeder's velocity * time
Since both distances are equal, we have the equation:
(1/2) * acceleration * time^2 = Speeder's velocity * time
Rearranging the equation, we get:
(1/2) * acceleration * time - Speeder's velocity * time = 0
Simplifying further:
(1/2) * acceleration - Speeder's velocity = 0
Substituting the given values:
(1/2) * 4.50 m/s^2 - 22 m/s = 0
Now, solve for time:
(1/2) * 4.50 m/s^2 * time - 22 m/s * time = 0
(1/2) * 4.50 m/s^2 * time = 22 m/s * time
(1/2) * 4.50 m/s^2 = 22 m/s
4.50 m/s^2 = 44 m/s
Now, divide both sides by 4.50 m/s^2 to find:
time = 11.11 s
Therefore, it takes approximately 11.11 seconds for the police car to catch the speeder.
(b) To find how far the two cars have traveled in this time, we can use the distance equation for each car.
Distance of the police car:
Distance = Initial velocity * time + (1/2) * acceleration * time^2
Distance = 0 * 11.11s + (1/2) * 4.50 m/s^2 * (11.11 s)^2
Simplify the equation to find the distance covered by the police car.
Distance of the speeder:
Distance = Speeder's velocity * time
Distance = 22 m/s * 11.11 s
Calculate the distance covered by the speeder.
(c) To find the velocity of the police car when it catches the speeder, we need to calculate the final velocity of the police car.
Final velocity of the police car:
Velocity = Initial velocity + acceleration * time
Velocity = 0 + 4.50 m/s^2 * 11.11 s
Compute the final velocity of the police car.
Please note that the numerical values used in this explanation are based on the information provided in the question, and the final answers may differ slightly depending on the precision of these values.
i cant seem to get this question though .
120 grams of an unknown metal at a temperature of 95 degrees celcuius is poured into 140 grams of aluminum calorimeter containing 220 grams of water. If the inital temperature of the water and calorimeter was 15 degree celcuius and the final temperature was 18 celcius, find the specfic heat of metal. the specfic heat of aluminum is .22 calories/gams Celcius. The specfic heat of water is 1.00 calorie /gram Celcius