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?

Respuesta :

Answer:

a) 19.1s

b) 599.4m

c) 62.8 m/s

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: [tex]s = v_0t = 31.4t m[/tex]

- The Officer: [tex]s = at^2/2 = 3.29t^2/2 = 1.645 t^2 m[/tex]

(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)

[tex]31.4t = 1.645t^2[/tex]

[tex]t = 31.4/1.645 = 19.1 s[/tex]

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

[tex]s = 1.645*19.1^2 = 599.4 m[/tex]

(c) His speed when he catches up would be

[tex]v = at = 3.29*19.1 = 62.8 m/s[/tex]