A motorist drives along a straight road at a constant speed of 16.5 m/s. Just as she passes a parked motorcycle police officer, the officer starts to accelerate at 3.50 m/s2 to overtake her. Assume that the officer maintains this acceleration.

(a) Determine the time it takes the police officer to reach the motorist.

(b) Find the speed of the officer as he overtakes the motorist.

(c) Find the total displacement of the officer as he overtakes the motorist.

Sm = distance travelled by motorist

Sp = distance travelled by officer
NOTE: at the time officer catches-up with motorist, Sm = Sp
t = time taken to catch-up with motorist
Ap = officer's acceleration
Up = officer's initial velocity = 0
Vp = officer's final velocity at the time he caught up with motorist

Sm = Umt
Sp = 1/2Apt2

Sm - Sp = 0 = 16.5t - 1/2x3.5t2

t(16.5 - 1.75t) = 0

t = 0 or t = 16.5/1.75 = 9.43s

Explanation: t = 0 is when the motorist passed the police car; and t = 9.43 is when the police car caught-up with the motorists.

(b) At the time officer caught up with motorist:
Up = 0, Ap = 3.5, t = 9.43 Vp = ?

v = u + at
Vp = Up + Apt
Vp = 0 + 3.5 x 9.43 = 33m/s

(c) Sp = 1/2(Apt^2) ===u = 0
s = 1/2 x 3.5 x 9.43^2 = 155.6m

To solve these problems, we need to use the equations of motion. In this case, we have a constant speed for the motorist, and the police officer is accelerating to catch up with her.

(a) To determine the time it takes for the police officer to reach the motorist, we can use the equation of motion:

\[v = u + at\]

Where:
v = final velocity (speed of the police officer)
u = initial velocity (starting speed of the police officer)
a = acceleration
t = time

Since the motorist is moving at a constant speed, her velocity (v) remains at 16.5 m/s, while the police officer starts from rest (u = 0) and accelerates at 3.50 m/s^2. Plugging these values into the equation, we get:

\[16.5 = 0 + 3.50t\]

Simplifying the equation, we have:

\[3.50t = 16.5\]

Solving for t:

\[t = \frac{16.5}{3.50}\]