Question

Challenge Problem (Extra Credit) Your good friend’s car (mentioned above) ignores a stop sign and enters an intersection with a speed v0 of [05] 31.4 m/s. He then continues with a constant velocity. A motorcycle traffic patrol officer is parked at the intersection. He discards his hot cocoa and doughnut and begins accelerating (at the instant the car passes him) with a constant acceleration of 3.29 m/s2.(a) How long (s) does it take the officer to catch the car?(b) How far (m) has the officer traveled when he catches up to the car?(c) How fast (m/s) is the officer going when he catches up to the car?

Answer 1

To catch the car, the police motorcycle needs approximately 19.09 seconds, travels a distance of 603.36 meters, and reaches a final velocity of 62.80 m/s, with all values rounded to two decimal places.

Step-by-step explanation:

To calculate the time t it takes for the motorcycle officer to catch up to the car, we need to set the equation for the displacement of both vehicles equal to each other and solve for t. The car travels with a constant velocity v0 of 31.4 m/s, so its displacement x after time t is simply v0 * t. The police motorcycle, however, starts from rest but accelerates at 3.29 m/s², so its displacement can be represented by the equation x = 0.5 * a * t² where a is the acceleration.

To find the time when the police motorcycle catches up to the car, we equate the two equations:
31.4 * t = 0.5 * 3.29 * t²

Solving for t, we get t = 2 * 31.4 / 3.29 = 19.09 seconds (when rounded to two decimal places).

The distance covered by the officer when he catches the car is found by substituting the time into the displacement equation for the police motorcycle:
0.5 * 3.29 * 19.09² = 603.36 meters (rounded to two decimal places).

The final velocity v of the officer can be calculated using the equation v = a * t:
3.29 * 19.09 = 62.80 m/s (rounded to two decimal places).

Answer 2

a) 19.1s

b) 599.4m

c) 62.8 m/s

Step-by-step explanation:

Suppose both the officer and your good friend start at the same position (the intersection point). The equation of motion for each of them are:

- Your friend:
`s = v_0t = 31.4t m`

- The Officer:
`s = at^2/2 = 3.29t^2/2 = 1.645 t^2 m`

(a) In order for the officer to catch him, they should both end up in the same position at the same time (other than t = 0)

`31.4t = 1.645t^2`

`t = 31.4/1.645 = 19.1 s`

(b) Within 19.1 s, the officer would have travelled a distance of

`s = 1.645*19.1^2 = 599.4 m`

(c) His speed when he catches up would be

`v = at = 3.29*19.1 = 62.8 m/s`