Answer:
The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.
Explanation:
Hi there!
The position of the patrol car at a time "t" can be calculated using this equation:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the patrol car at a time "t"
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:
x = x0 + v · t
Where "v" is the constant velocity.
First, let´s convert the velocity units into m/s:
140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s
105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s
We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.
When the patrol car catches the speeder, the position of both cars is the same:
position of the patrol car = position of the speeder
x0 + v0 · t + 1/2 · a · t² = x0 + v · t
if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:
x = v · t
x = 38.9 m/s · 1 s = 38.9 m
When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:
x0 + v0 · t + 1/2 · a · t² = x0 + v · t
0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t
Let´s agrupate terms and equalize to zero:
-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0
-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0
Solving the quadratic equation for t using the quadratic formula:
t = 8.24 s (the other solution is discarded because it is negative)
The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.
