The police car immediately begins to accelerate as the \"speeder\" passes the st

Question

The police car immediately begins to accelerate as the "speeder" passes the stationary police car. The police car's acceleration is 2.9 m/s^2. By the way 90mph is (almost if not equal to) 40m/s.

1.)How much time passes until the police car has the same speed as the speeder? A=14 seconds
2.)How much time passes until the police car has caught up to the speeder? A=28 seconds
3.)How fast is the police car moving when it has caught up to the speeder? A=80m/s
4.)How far from the original position is the police car when it has caught up to the speeder? A=1,103 or 1100 meters

Explanation / Answer

(a) For the police officer: d = ½ . a . t² For the speeder: d = v . t The police officer can be considered to have passed the speeder when they have travelled an equal distance, thus, we substitute the second equation into the first, giving us. v . t = ½ . a . t² Solving this for t gives us: v . t = ½ . a . t² (÷ t) v = ½ . a . t (÷ a) v/a = ½ . t (× 2) 2.v/a = t Substituting in the values gives us: t = 2.v/a t = 2 . 30 / 2.44 t = 24.590 s (b) Now we know how long it took. v = d/t d = v.t d = 30 . 24.590 s d = 737.705 m

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