A motorist drives along a straight road at a constant speed of 15.0 m/s. just as she passes a parked police officer,the officer starts to accelerate at 2.00 m/s^2 to overtake her. assuming the officer maintains this acceleration, a) determine the time it takes the police officer to reach the motorist. find b) the speed and c) the total displacement of the officer as he overtakes the motorist.

A. 15

B. 30
C. 225

physics

To solve this problem, we need to analyze the motion of both the motorist and the police officer. We'll use the equations of motion to find the answers. Here's how you can do it:

a) Determine the time it takes the police officer to reach the motorist:
Let's assume that the police officer catches up with the motorist after time 't'. The motorist is traveling at a constant speed of 15.0 m/s. Assuming the initial position of the police officer is zero, and the motorist is some distance ahead of the officer. We can use the equation:

Distance covered by the motorist = Distance covered by the police officer

The distance covered by the motorist is given by: Distance_motorist = speed x time = 15.0t.

The distance covered by the police officer is given by the equation of motion: Distance_officer = (initial velocity x time) + (0.5 x acceleration x time^2). Here, the initial velocity of the police officer is zero.

Setting the two distances equal to each other:
15.0t = 0.5 x 2.00 x t^2

Simplifying and rearranging the equation:
0.5 x 2.00 x t^2 - 15.0t = 0

Now we have a quadratic equation. We can solve it using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values:
t = (-(-15.0) ± √((-15.0)^2 - 4 x 0.5 x 2.00 x 0)) / (2 x 0.5 x 2.00)

Simplifying further:
t = (15.0 ± √(225 - 0)) / (2.00)

t = (15.0 ± √225) / 2.00

t = (15.0 ± 15.0) / 2.00

Now we have two solutions for t. Since time cannot be negative, we only consider the positive value:

t = (15.0 + 15.0) / 2.00
t = 30.0 / 2.0
t = 15.0 seconds

Therefore, it takes the police officer 15.0 seconds to reach the motorist.

b) Find the speed of the police officer when he overtakes the motorist:
We can use the equation of motion to find the speed of the police officer when he overtakes the motorist. The equation is: Speed = initial velocity + acceleration x time.

Here, the initial velocity of the police officer is zero, and we found that the time is 15.0 seconds.

Speed = 0 + 2.00 x 15.0
Speed = 30.0 m/s

Therefore, the speed of the police officer when he overtakes the motorist is 30.0 m/s.

c) Find the total displacement of the police officer as he overtakes the motorist:
To find the displacement, we can use the equation: Displacement = initial velocity x time + 0.5 x acceleration x time^2.

Here, the initial velocity of the police officer is zero, and we found that the time is 15.0 seconds.

Displacement = 0 + 0.5 x 2.00 x (15.0)^2
Displacement = 0.5 x 2.00 x 225.0
Displacement = 225.0 m

Therefore, the total displacement of the police officer as he overtakes the motorist is 225.0 meters.

Let V = motorist speed

a = officer acceleration
a) V*t = (1/2)a t^2
Solve for t when they are at same place
b) a*t = cop speed at time t
c) (1/2) at^2 = cop displacement at time t

= (1/2)(2V/a)^2*a = 2 V^2/a