A speeder traveling at 30 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 2.4 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder?
We want the distances to be equal, so
30t = 1/2 * 2.4 t^2
And now solve for t.
driver goes 30 t
cop goes .5 a t^2 = 1.2 t^2
so
30 t = 1.2 t^2
t = 30/1.2 seconds
or of course they are also together at t = 0 :)
To determine how long it takes for the policeman to catch the speeder, we need to find the time it takes for their velocities to become equal. Let's break down the problem into steps:
Step 1: Find the distance traveled by the speeder during the time taken by the policeman to catch up.
First, we need to find the distance the speeder travels before the policeman catches up. Since we don't know the total time, we'll have to calculate it later.
Step 2: Set up equations for the motion of the speeder and the policeman.
Let's assign some variables:
- t: time taken for the policeman to catch the speeder (this is what we need to find)
- d: distance traveled by the speeder during time t
For the speeder:
The velocity of the speeder (v_speeder) is given as 30 m/s.
For the policeman:
The velocity of the policeman (v_policeman) is initially 0 m/s since he is at rest. However, since he is accelerating at a rate of 2.4 m/s^2, his velocity can be given as:
v_policeman = 2.4 * t
Step 3: Equate the distances traveled by the speeder and the policeman.
The distance traveled by an object can be calculated by multiplying its velocity by time. Equating the distances of the speeder and the policeman at the moment they are at the same position, we have:
d = v_speeder * t
d = (30 * t)
Setting the distances equal, we have:
(30 * t) = (2.4 * t)
Step 4: Solve the equation to find t.
Let's solve for t:
30t = 2.4t
27.6t = 0
t = 0
Our equation gives us t = 0. However, this doesn't make sense in our scenario, as the policeman needs time to catch the speeder. So we need to reconsider our approach.
Step 5: Finding the correct equation and solving for t.
Since the policeman is accelerating, the equation needs some modification. We'll use an equation of motion that involves displacement, initial velocity, acceleration, and time:
d = v_0 * t + (1/2) * a * t^2
For the policeman, the initial velocity (v_0) is 0 m/s, the acceleration (a) is 2.4 m/s^2, and the displacement (d) is the distance traveled by the speeder.
Setting the two equations equal, we have:
(30 * t) = (0 * t) + (1/2) * (2.4) * t^2
Simplifying this equation, we get:
30t = 1.2t^2
Dividing both sides of the equation by t, we get:
30 = 1.2t
Dividing both sides by 1.2, we find:
t ≈ 25
Therefore, to the nearest tenth of a second, it takes the policeman approximately 25 seconds to catch the speeder.