Question

Physics with Calc I

A motorist drives along a straight road at a constant speed of 15.5 m/s. Just as she passes a parked motorcycle police officer, the officer starts to accelerate at 3.00 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.

(d) Draw position, velocity, and acceleration graphs for the motorist and officer. You should have 3 graphs, one for position (with both motorist and officer on same graph), one for velocity (with both motorist and officer on same graph), and one for acceleration (same as before).

Answer 1

(a) Take the instant when the motorist passes the cop as the origin of an x-axis; the time starts here and the position of both vehicles is zero. The displacement of the motorist is:

Δx = (15.5 m/s)t----------------->(1)
The displacement of the cop is:
Δx = 0.5(3.00m/s²)t² = (1.5 m/s²)t²--------------------->(2)

Since when the cop catches the car, their displacements will be equal. Therefore, we can set (1) equal to (2) and solve for t:

(1.5 m/s²)t² = (15.5 m/s)t
t = 15.5 m/s / 1.5 m/s² = 10.33 s

(b) When he overtakes the motorist, he will have travelled for 10.33 s, so his speed will be:
v = v₀ + at
= 0 + (3.00 m/s²)(10.33 s) = 30.99 m/s

(c) His displacement is found from (2):
Δx = 0.5(3.00m/s²)t²
= 0.5(3.00m/s²)(10.33 s)² = 160.06 m

Hope this helps.