A speeder passes a parked car at 35.0 m/s. The police car starts from rest with a uniform acceleration of 2.41 m/s^2.
A.) How much time passes before the speeder is overtaken by the police car?
B.) How far does the speeder get before being overtaken by the police car?
A.) How much time passes before the speeder is overtaken by the police car?
Well, it seems like the police car is on the case! Let's calculate how much time it takes for the police car to catch up to the speedy speeder.
We can use the equation of motion: v = u + at
Where:
v = final velocity (speeder's velocity)
u = initial velocity (police car's velocity, which is zero because it starts from rest)
a = acceleration (police car's acceleration)
t = time
We know that the speeder's velocity is 35.0 m/s and the police car's acceleration is 2.41 m/s^2. So, let's rearrange the equation to solve for time:
35.0 m/s = 0 + 2.41 m/s^2 × t
Now, let's solve for t:
t = 35.0 m/s ÷ (2.41 m/s^2)
t ≈ 14.52 seconds
So, it takes approximately 14.52 seconds for the police car to catch up with the speeder.
B.) How far does the speeder get before being overtaken by the police car?
To find out how far the speeder gets before being overtaken, we need to use the equation of motion: s = ut + ½at²
Where:
s = distance
u = initial velocity (police car's velocity, which is zero because it starts from rest)
t = time (we already calculated this to be approximately 14.52 seconds)
a = acceleration (police car's acceleration)
Since the police car starts from rest, the initial velocity is zero, so the equation simplifies to:
s = 0 × 14.52 + ½ × 2.41 m/s^2 × (14.52 s)²
Now, let's calculate the distance:
s ≈ 0 + 0.5 × 2.41 m/s^2 × 210.7504 s²
s ≈ 254.2 meters
So, the speeder gets approximately 254.2 meters before being overtaken by the police car.
To find the time it takes for the police car to overtake the speeder, we need to determine when their positions are equal.
Let's start by finding the position of the police car at any given time t. We can use the kinematic equation:
s_police = 0 + (1/2) * a_police * t^2
where s_police is the position of the police car, a_police is the acceleration of the police car, and t is the time.
Since the police car starts from rest, its initial velocity is 0, and we can substitute the value of the acceleration:
s_police = (1/2) * 2.41 * t^2
Now let's find the position of the speeder at any given time t. The speeder is traveling at a constant velocity, so we can use the equation:
s_speeder = v_speeder * t
where s_speeder is the position of the speeder and v_speeder is the velocity of the speeder.
Given that the speeder passes the parked car at 35.0 m/s, the equation becomes:
s_speeder = 35.0 * t
Now, since the police car overtakes the speeder, their positions are equal:
s_police = s_speeder
Substituting the equations for s_police and s_speeder:
(1/2) * 2.41 * t^2 = 35.0 * t
Simplifying the equation, we get a quadratic equation:
1.205 * t^2 - 35.0 * t = 0
To solve this equation, we can apply the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1.205, b = -35.0, and c = 0. Plugging in these values:
t = (-(-35.0) ± √((-35.0)^2 - 4 * 1.205 * 0)) / (2 * 1.205)
Simplifying further:
t = (35.0 ± √(1225.0)) / 2.41
Now, we have two possible solutions for t:
t1 = (35.0 + √(1225.0)) / 2.41
t2 = (35.0 - √(1225.0)) / 2.41
Since time cannot be negative, we can discard t2. Therefore, the time it takes for the police car to overtake the speeder is:
t = (35.0 + √(1225.0)) / 2.41
To find the distance the speeder gets before being overtaken by the police car, we can substitute this value of t into the equation for s_speeder:
s_speeder = 35.0 * t
Calculating this, we can find the distance the speeder gets before being overtaken by the police car.