A speeder passes a parked police car at a
constant speed of 31.8 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.74 m/s
2
.
How much time passes before the speeder
is overtaken by the police car?
vt= at²/2
t=2v/a=2•31.2/2.74= 22.8 s
To find the time it takes for the police car to overtake the speeder, we can use the equations of motion. Let's break down the problem.
We are given:
- Speed of the speeder, v1 = 31.8 m/s
- Acceleration of the police car, a2 = 2.74 m/s^2
- Initial velocity of the police car, u2 = 0 m/s
We need to find:
- Time taken for the police car to overtake the speeder, t.
First, let's find the equation that relates the distance traveled by both the speeder and the police car.
The distance traveled by the speeder, S1, can be calculated using the equation:
S1 = v1 * t
The distance traveled by the police car, S2, can be calculated using the equation of motion:
S2 = u2 * t + (1/2) * a2 * t^2
Since both the speeder and the police car cover the same distance when the police car overtakes the speeder, we can set S1 = S2.
So, we have the equation:
v1 * t = (1/2) * a2 * t^2
Rearranging the equation, we get:
(1/2) * a2 * t^2 - v1 * t = 0
This is a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values into the quadratic formula, we get:
t = (-(-v1) ± √((-v1)^2 - 4*(1/2)*a2*0)) / (2*(1/2)*a2)
Simplifying further, we get:
t = (v1 ± √(v1^2)) / a2
Since time cannot be negative, we take the positive solution:
t = (v1 + √(v1^2)) / a2
Plugging in the values, we have:
t = (31.8 + √(31.8^2)) / 2.74
Evaluating the expression, the time taken for the police car to overtake the speeder is approximately 19.32 seconds.