A speeding motorcyclist is traveling at a constant speed of 31.1 m/s when he passes a police car parked on the side of the road. The radar, positioned in the police car's rear window, measures the speed of the motorcycle. At the instant the motorcycle passes the police car, the police officer starts to chase the motorcyclist with a constant acceleration of 4.11 m/s2.
a) How long will it take the police officer to catch the motorcyclist?
b) What is the speed of the police car when it catches up to the motorcycle?
c) How far will the police car be from its original position?
a) How long will it take the police officer to catch the motorcyclist?
Well, catching a motorcyclist is no easy task, especially if they're speeding! But with a constant acceleration of 4.11 m/s², our determined police officer will eventually catch up. To find out how long it will take, we can use the following equation:
Δx = v₀t + 0.5at²
Since the motorcycle is traveling at a constant speed of 31.1 m/s, its initial velocity v₀ is 31.1 m/s. The police officer starts from rest, so their initial velocity is 0 m/s. The acceleration of the police car is 4.11 m/s². We need to solve for t, the time it takes for the police officer to catch the motorcyclist.
When the police officer catches the motorcyclist, their positions will be the same. Therefore, we can set Δx equal to zero:
0 = (31.1 m/s)(t) + 0.5(4.11 m/s²)(t²)
Simplifying the equation gives us a quadratic equation:
0 = (0.5(4.11 m/s²))(t²) + (31.1 m/s)(t)
Using the quadratic formula to solve for t, we find:
t = (-31.1 ± √(31.1² - 4(0.5(4.11))(0))) / (2(0.5(4.11)))
After some math, we get two possible solutions for t: -6.81s and 4.37s. Since time cannot be negative, we take the positive value, which gives us t ≈ 4.37 seconds. So it will take the police officer roughly 4.37 seconds to catch the motorcyclist.
b) What is the speed of the police car when it catches up to the motorcycle?
When the police car catches up to the motorcycle, their velocities will be the same. We need to find the velocity of the police car at that moment.
Using the formula:
v = v₀ + at
Since the police car starts from rest (v₀ = 0 m/s) and accelerates at a constant rate of 4.11 m/s² for 4.37 seconds, we can plug in the values:
v = (0 m/s) + (4.11 m/s²)(4.37 s)
Calculating this gives us:
v ≈ 17.95 m/s
So, the speed of the police car when it catches up to the motorcycle is approximately 17.95 m/s.
c) How far will the police car be from its original position?
To determine how far the police car will be from its original position, we can use the following equation:
Δx = v₀t + 0.5at²
Since the initial velocity of the police car is 0 m/s and the time it takes to catch the motorcyclist is 4.37 seconds, we can plug in the values:
Δx = (0 m/s)(4.37 s) + 0.5(4.11 m/s²)(4.37 s)²
Simplifying this equation gives us:
Δx = (0 m) + (0.5 m/s²)(9.01 s²)
Multiplying and calculating further yields:
Δx ≈ 19.55 m
So, the police car will be approximately 19.55 meters from its original position.
To solve this problem, we need to use equations of motion.
a) To find the time it takes for the police officer to catch the motorcyclist, we can use the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
Since the motorcyclist and the police officer start at the same point, we can equate their distances:
31.1t + (1/2)(4.11)(t^2) = 0
Rearranging the equation:
2.055t^2 + 31.1t = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 2.055, b = 31.1, and c = 0
t = (-31.1 ± √(31.1^2 - 4(2.055)(0))) / (2(2.055))
t ≈ 3.01 seconds
Therefore, it will take the police officer approximately 3.01 seconds to catch the motorcyclist.
b) To find the speed of the police car when it catches up to the motorcycle, we can use the equation:
final velocity = initial velocity + acceleration * time
v = 31.1 + 4.11 * 3.01
v ≈ 44.25 m/s
Therefore, the speed of the police car when it catches up to the motorcycle is approximately 44.25 m/s.
c) To find the distance the police car will be from its original position, we can use the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
distance = 31.1 * 3.01 + (1/2) * 4.11 * (3.01^2)
distance ≈ 139.807 meters
Therefore, the police car will be approximately 139.807 meters from its original position.
To solve this problem, we can use the equations of motion to determine the time it takes for the police officer to catch up to the motorcyclist, the speed of the police car when it catches up, and the distance the police car travels from its original position.
a) To find the time it will take for the police officer to catch the motorcyclist, we need to determine the time it takes for both the motorcyclist and the police car to cover the same distance.
The motorcyclist travels at a constant speed of 31.1 m/s, so the time needed for the motorcyclist to cover the distance is given by t = d/v, where d is the distance and v is the constant speed. However, since the problem does not provide the distance the motorcyclist travels, we need to find it.
We can use the equation of motion for uniformly accelerated motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
For the motorcyclist, the initial velocity (u) is 31.1 m/s, the final velocity (v) is 31.1 m/s, and the acceleration (a) is 0 m/s^2 because the motorcyclist travels at a constant speed. Hence, we can rewrite the equation as 31.1^2 = 31.1^2 + 2(0)s.
Simplifying the equation, we get 0s = 0.
Since the displacement (s) is zero, it means that the motorcyclist does not change its position. Therefore, the time it takes for the motorcyclist to reach the position of the police car is zero.
Now, for the police car:
Using the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
The initial velocity (u) of the police car is zero because it starts from rest, the final velocity (v) is the same as the motorcyclist's speed at 31.1 m/s, and the acceleration (a) is given as 4.11 m/s^2. We can rewrite the equation as 31.1 = 0 + 4.11t.
Simplifying the equation, we find t = 31.1/4.11 = 7.56 seconds.
Therefore, it will take the police officer 7.56 seconds to catch up to the motorcyclist.
b) Now, to find the speed of the police car when it catches up to the motorcycle, we can use the equation of motion: v = u + at.
Plugging in the values, the initial velocity (u) of the police car is zero, the acceleration (a) is 4.11 m/s^2, and the time (t) is 7.56 seconds. We can rewrite the equation as v = 0 + (4.11)(7.56).
Simplifying the equation, we find v = 31.10 m/s.
Therefore, the speed of the police car when it catches up to the motorcycle is 31.10 m/s.
c) To find the distance the police car travels from its original position, we can use the equation of motion: s = ut + (1/2)at^2.
The initial velocity (u) of the police car is zero, the acceleration (a) is given as 4.11 m/s^2, and the time (t) is 7.56 seconds. Plugging in these values, we have s = 0 + (1/2)(4.11)(7.56)^2.
Simplifying the equation, we find s = 119.2 meters.
Therefore, the police car will be approximately 119.2 meters away from its original position when it catches up to the motorcycle.
(a)
31.1t = 1/2*4.11t^2
t = 15.134 seconds
Now use that t for parts (b) and (c)