A motorist drives along a straight road at a constant speed of 15.0 m/s. Just as she passes a parked motorcycle police officer, the officer starts to accelerate at 2.00 m/s2 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.
did no one solve this :((((
To solve this problem, we'll use the equations of motion, specifically the equation of motion for uniformly accelerated linear motion. This equation relates the displacement (Δx), initial velocity (v₀), final velocity (v), acceleration (a), and time (t):
Δx = v₀t + (1/2)at²
(a) To find the time it takes for the police officer to reach the motorist, we need to determine the time when the displacement of the officer (Δx) equals the distance traveled by the motorist. Since the motorist is traveling at a constant speed of 15.0 m/s, the displacement of the motorist in time t is given by:
Displacement of motorist = v_motorist * t
Since the motorist and the officer start at the same location, we can set the displacement of motorist equal to the displacement of the officer:
Δx = v_motorist * t
Now, let's calculate the displacement of the officer. The officer starts from rest and accelerates at 2.00 m/s², so the equation of motion becomes:
Δx_officer = (1/2)at²
Plugging in the values:
Δx_officer = (1/2) * 2.00 * t²
Δx_officer = t²
Setting Δx_officer equal to v_motorist * t:
t² = v_motorist * t
Since the motorist is traveling at a constant speed of 15.0 m/s, we substitute this:
t² = 15.0 * t
Rearranging the equation:
t² - 15t = 0
Factoring out a t:
t(t - 15) = 0
From this equation, we have two possible solutions for t: t = 0 and t = 15. The first solution t = 0 implies that the officer has not yet started moving, so we discard it. Therefore, the time it takes for the police officer to reach the motorist is 15 seconds.
(b) To find the speed of the police officer at the moment of overtaking, we can use the equation:
v = v₀ + at
Substituting the known values:
v = 0 + 2.00 * 15
v = 30.0 m/s
So, the speed of the police officer at the moment of overtaking is 30.0 m/s.
(c) The total displacement of the officer as he overtakes the motorist is given by:
Δx_officer = (1/2)at²
Substituting the known values:
Δx_officer = (1/2) * 2.00 * (15)²
Δx_officer = 225 m
Therefore, the total displacement of the police officer as he overtakes the motorist is 225 meters.