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Unformatted text preview: t o ) the gap between the cars is 10.6 m. The speed of the police car at t o is 30.56 5(2.4) = 18.56 m/s. Collision occurs at time t when x you = x police (we choose coordinates such that your position is x = 0 and the police cars position is x = 10.6 m at t o ). Eq. 215 becomes, for each car: x police 10.6 = 18.56( t t o ) 1 2 (5)( t t o ) 2 x you = 30.56( t t o ) 1 2 (5)( t t o ) 2 . Subtracting equations, we find 10.6 = (30.56 18.56)( t t o ) 0.883 s = t t o . At that time your speed is 30.56 + a ( t t o ) = 30.56 5(0.883) 26 m/s (or 94 km/h)....
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 Spring '06
 Stoler
 Physics, Light

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