A speeder passes a parked police car at a
constant speed of 25.8 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.68 m/s^2.
How much time passes before the speeder
is overtaken by the police car?
Answer in units of s
How far does the speeder travel before being
overtaken by the police car?
Answer in units of m.
s=2u^2/a
s=2*25.8^2/2.68
s=496.75
s=ut
t=s/u
t=496.75/25.8
t=19.25sec
The police car catches up 2.68 m/s for every elapsed second. So to catch up 25.8 m/s, it will take
t=25.8 m/s^2 / 2.68 m/s
= 9.63 s
Since the speeder travels at a uniform speed of 25.8 m/s, the distance travelled can be calculated by
Distance = speed * time.
To find the time it takes for the police car to overtake the speeder, we need to determine when both vehicles will have traveled the same distance.
Let's consider the time it takes for the police car to catch up to the speeder as "t". During this time, the speeder will continue moving at a constant speed of 25.8 m/s.
We can start by calculating the distance travelled by the speeder during time "t" using the equation:
Distance = Speed * Time
Distance traveled by the speeder = 25.8 m/s * t
Now let's consider the distance travelled by the police car during time "t". Since the police car starts from rest, we can use the equation for distance traveled under constant acceleration:
Distance = Initial velocity * Time + 0.5 * Acceleration * Time^2
The initial velocity of the police car is 0 m/s. Substituting the values into the equation gives us:
Distance traveled by the police car = 0.5 * 2.68 m/s^2 * t^2
Now we can set the two distances equal and solve for "t":
25.8 m/s * t = 0.5 * 2.68 m/s^2 * t^2
Divide both sides of the equation by "t" to simplify:
25.8 m/s = 0.5 * 2.68 m/s^2 * t
Multiply both sides by 2 to get rid of the fraction:
51.6 m/s = 2.68 m/s^2 * t
Divide both sides by 2.68 m/s^2 to solve for "t":
t = 51.6 m/s / 2.68 m/s^2
t ≈ 19.25 s
Therefore, it takes approximately 19.25 seconds for the police car to overtake the speeder.
Now, let's calculate the distance traveled by the speeder during that time:
Distance traveled by the speeder = 25.8 m/s * t
Distance traveled by the speeder = 25.8 m/s * 19.25 s
Distance traveled by the speeder ≈ 496.95 m
Therefore, the speeder will travel approximately 496.95 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 the position or distance traveled by both vehicles is the same.
Let's break down the problem and start by finding the time it takes for the police car to catch up to the speeder.
Step 1: Determine the equations relating distance, initial velocity, acceleration, and time for each vehicle:
For the speeder:
Distance traveled by the speeder = Initial velocity * time + 0.5 * 0 * time^2 (since the speeder maintains a constant velocity)
We can simplify this equation to:
Distance traveled by the speeder = Initial velocity * time
For the police car:
Distance traveled by the police car = 0 * time + 0.5 * acceleration * time^2
We can simplify this equation to:
Distance traveled by the police car = 0.5 * acceleration * time^2
Step 2: Set the two distance equations equal to each other:
Initial velocity * time = 0.5 * acceleration * time^2
Step 3: Solve the equation for time:
Initial velocity * time = 0.5 * acceleration * time^2
Rearranging the equation:
0.5 * acceleration * time^2 - Initial velocity * time = 0
This is a quadratic equation with the form ax^2 + bx + c = 0.
Here, a = 0.5 * acceleration, b = -Initial velocity, and c = 0.
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a,
we can substitute the values and solve for time.
time = (-(-Initial velocity) ± √((-Initial velocity)^2 - 4 * 0.5 * acceleration * 0)) / (2 * 0.5 * acceleration)
time = (Initial velocity ± √(Initial velocity^2)) / acceleration
Step 4: Calculate time:
time = (Initial velocity + √(Initial velocity^2)) / acceleration
Substituting the values:
time = (25.8 m/s + √((25.8 m/s)^2)) / 2.68 m/s^2
Now, let's calculate the time:
time = (25.8 m/s + √(665.64 m^2/s^2)) / 2.68 m/s^2
time = (25.8 m/s + 25.8 m/s) / 2.68 m/s^2
time = 51.6 m/s / 2.68 m/s^2
time ≈ 19.254 seconds
Therefore, it takes approximately 19.254 seconds for the police car to overtake the speeder.
To find how far the speeder travels before being overtaken by the police car, we can substitute the found time back into the distance formula for the speeder.
Distance traveled by the speeder = Initial velocity * time
Distance traveled by the speeder = 25.8 m/s * 19.254 s
Distance traveled by the speeder ≈ 497.8572 meters
Therefore, the speeder travels approximately 497.8572 meters before being overtaken by the police car.