A speeder passes a parked police car at a
constant speed of 20.1 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.65 m/s
How much time passes before the speeder
is overtaken by the police car?
Answer in units of s
d = u t for car
d = 20.1 t
d = (1/2) a t^2 for cruiser
d = (1/2) 2.65 t^2 = 1.325 t^2
so
1.325 t^2 - 20.1 t = 0
t (1.325 t - 20.1) = 0
they are together at t = 0
and again at
t = 20.1/1.325 = 15.2 seconds
To find the time it takes for the police car to overtake the speeder, we can use the equations of motion. Firstly, we need to determine the distance traveled by both the speeder and the police car at the point of overtaking.
Let's assume that the time taken for the police car to overtake the speeder is t.
For the speeder:
The speed of the speeder is constant at 20.1 m/s. So, the distance traveled by the speeder can be represented as: distance = speed × time. Therefore, the distance traveled by the speeder is given by:
Distance of the speeder = 20.1 m/s × t
For the police car:
The police car starts from rest and has a uniform acceleration of 2.65 m/s². Using the equation of motion, s = ut + (1/2)at² (where s is distance, u is initial velocity, a is acceleration, and t is time), we can find the distance traveled by the police car.
The initial velocity of the police car is 0 m/s (since it starts from rest), and the acceleration is given as 2.65 m/s². Therefore, the distance traveled by the police car is given by:
distance = (1/2) × acceleration × time²
Now, for the point of overtaking, the distance traveled by the speeder and the police car will be the same. Therefore, we can equate the two equations:
20.1 m/s × t = (1/2) × 2.65 m/s² × t²
Simplifying the equation:
20.1 m/s × t = 1.325 m/s² × t²
Rearranging the equation:
1.325 m/s² × t² - 20.1 m/s × t = 0
Now, we can solve this quadratic equation to find the value of t.