Posted by **Becky** on Sunday, October 11, 2009 at 11:33pm.

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.

- Physics -
**HEllo**, Friday, September 9, 2016 at 1:04pm
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

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