posted by Becky .
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