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]