Question

A tourist is driving a car along a straight road at a constant speed of 22.5 m/s. Just as she passes a law officer parked along the side of the road, the officer starts to accelerate 2.50 m/s² to overtake her. Assume that the officer maintains this acceleration.

(a) Determine the time it takes the officer to reach the tourist (in s).

18 s

(b) Find the speed of the officer as he overtakes the tourist (in m/s).

  45 m/s

(c) Find the total displacement of the officer as he overtakes the tourist (in m).

  405 m

(d) What If? Determine the answers to parts (a) through (c) of the problem, if instead of accelerating at the same instant the car passes him, the officer instead begins his acceleration 1.00 s after the car passes him. (Enter the time in s, speed in m/s, and displacement in m.)

time _____s

speed ____m/s

displacement _____m

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 = (22.5m/s)t----------------->(1)

The displacement of the cop is:

Δx = 0.5(2.5m/s²)t²

= (1.25m/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:

(1.25m/s²)t² = (22.5m/s)t

t = 18 s

(b) When he overtakes the motorist, he will have travelled for 12.5s, so his speed will be:

v = v₀ + at

= 0 + (2.5m/s²)(18s)

= 45 m/s

(c) His displacement is found from (2):

Δx = 0.5(2.5m/s²)t²

= 0.5(2.5m/s²)(18s)²

= 405m

(d) This time the motorist travels an extra distance in that extra 1 s.

Thus, the displacement of the motorist is:

Δx = 22.5 + (22.5m/s)t

Equating:

(1.25m/s²)t² = 22.5 + (22.5m/s)t

=> 1.25t^2 – 22.5t – 22.5 = 0

=> t = (22.5 + sqrt(22.5^2 – 4*1.25*(-22.5)))/(2*1.25)

= 18.94 s

v = v₀ + at

= 0 + (2.5m/s²)(18.94s)

= 47.37 m/s

Δx = 0.5(2.5m/s²)t²

= 0.5(2.5m/s²)(18.94s)²

= 448.87 m

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