A car is speeding at 25 m/s in a school zone. A police car starts from rest just as the speeder passes and accelerates at a constant rate of 5 m/s2. How fast is the police car traveling when it is 25 m behind the speeding car?
I honestly don't know where to start. We don't know the final velocity (we're trying to find it, duh) and since we don't have the time, I don't know what to do.
Thanks.
you know the distance both cars went in the same time
d1=25*t
d1-25=1/2 5(t^2) then
25t-25=2.5 t^2
solve for time t, then how fast is ...
v=5t
Well, well, well, it seems like we have a classic case of a speeder on our hands! Let's see if we can catch that speedy culprit.
To find the velocity of the police car when it's 25 m behind the speeding car, we can use a little bit of kinematics. Don't worry, I'll explain it in a way that won't make your brain feel like it's going in circles!
We know that the speeder is going at a constant velocity of 25 m/s, and the police car starts from rest. So, let's first find out how much time it takes for the police car to catch up to the speeder.
Using the kinematic equation:
\(v_f = v_i + at\)
Where:
\(v_f\) is the final velocity (we want to find this),
\(v_i\) is the initial velocity of the police car (which is 0 m/s since it starts from rest),
\(a\) is the acceleration of the police car (which is 5 m/s²), and
\(t\) is the time it takes for the police car to catch up to the speeder.
Now, since we don't know the time, we can use another equation to find it. The equation relates displacement, initial velocity, time, and acceleration:
\(d = v_it + \frac{1}{2}at^2\)
In this equation:
\(d\) is the displacement of the police car (which is 25 m, as it needs to be 25 m behind),
\(v_i\) is the initial velocity of the police car (which is 0 m/s),
\(a\) is the acceleration of the police car (which is 5 m/s²), and
\(t\) is the time it takes for the police car to catch up.
Now, we can plug in the numbers and solve both equations simultaneously to find the time it takes for the police car to catch up with the speeder.
Once you've found the time, you can substitute it back into the first equation \(v_f = v_i + at\) to determine the velocity of the police car when it catches up with the speeder.
Now, remember, don't get too carried away with your joy of catching the speeder that you forget to give them a proper lecture! Safety first, my friend.
To solve this problem, we can use the equations of motion. Let's break down the problem step-by-step:
Step 1: Determine the initial velocity of the police car.
In the problem, it states that the police car starts from rest. Therefore, the initial velocity of the police car, u₁ = 0 m/s.
Step 2: Determine the acceleration of the police car.
In the problem, it states that the police car accelerates at a constant rate of 5 m/s². Therefore, the acceleration, a = 5 m/s².
Step 3: Determine the displacement of the police car.
The problem states that the police car needs to be 25 m behind the speeding car. Therefore, the displacement, s = -25 m (negative sign indicates that the police car is behind the speeding car).
Step 4: Determine the displacement traveled by the speeding car.
Since the police car starts from rest, the time it takes for the police car to reach the same displacement as the speeding car is the same as the time taken by the speeding car to travel that distance. Therefore, the displacement traveled by the speeding car can be determined using the equation:
s₁ = u₁t + (1/2)at²
Since the initial velocity of the speeding car, u₁ = 25 m/s, and the time taken for both cars to reach the same displacement is the same, we can simplify the equation to:
s₁ = 25t + (1/2)(5)t²
Step 5: Determine the time taken by both cars to travel the respective distances.
The problem states that both cars reach the same displacement. Therefore, we can equate the displacements of the police car and the speeding car.
-25 = 25t + (1/2)(5)t²
Step 6: Solve the quadratic equation and determine the time taken.
Rearranging the equation, we get:
(1/2)(5)t² + 25t + 25 = 0
Solve this quadratic equation using a suitable method (factoring, quadratic formula, etc.) to find the value(s) of t.
Step 7: Determine the final velocity of the police car.
Once you find the value of t, substitute it into the equation for the displacement of the police car to find the final velocity, v.
v = u₁ + at
Substitute the values of u₁ (0 m/s), a (5 m/s²) and t (obtained from the previous step) to find the final velocity of the police car when it is 25 m behind the speeding car.
To solve this problem, we can use a kinematic equation that relates acceleration, initial velocity, final velocity, and displacement. The equation we will use is:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
Let's break down the given information:
Initial velocity of the car (speeding car): u = 25 m/s
Acceleration of the police car: a = 5 m/s^2
Displacement between the cars: s = 25 m
We need to find the final velocity of the police car (v) when it is 25 m behind the speeding car.
Now, let's plug in the values in the equation:
v^2 = (25 m/s)^2 + 2 * 5 m/s^2 * 25 m
Simplifying:
v^2 = 625 m^2/s^2 + 250 m^2/s^2
v^2 = 875 m^2/s^2
To find v, we take the square root of both sides:
v = √(875 m^2/s^2)
Now, if we calculate the square root:
v ≈ 29.5 m/s
Therefore, the police car is traveling at approximately 29.5 m/s when it is 25 m behind the speeding car.