# ch02-p043 - t o the gap between the cars is 10.6 m The...

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43. (a) Note that 110 km/h is equivalent to 30.56 m/s. During a two second interval, you travel 61.11 m. The decelerating police car travels (using Eq. 2-15) 51.11 m. In light of the fact that the initial “gap” between cars was 25 m, this means the gap has narrowed by 10.0 m – that is, to a distance of 15.0 m between cars. (b) First, we add 0.4 s to the considerations of part (a). During a 2.4 s interval, you travel 73.33 m. The decelerating police car travels (using Eq. 2-15) 58.93 m during that time. The initial distance between cars of 25 m has therefore narrowed by 14.4 m. Thus, at the start of your braking (call it
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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 car’s position is x = 10.6 m at t o ). Eq. 2-15 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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## This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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