ch02-p031 - 31. We assume the periods of acceleration...

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31. We assume the periods of acceleration (duration t 1 ) and deceleration (duration t 2 ) are periods of constant a so that Table 2-1 can be used. Taking the direction of motion to be + x then a 1 = +1.22 m/s 2 and a 2 = –1.22 m/s 2 . We use SI units so the velocity at t = t 1 is v = 305/60 = 5.08 m/s. (a) We denote Δ x as the distance moved during t 1 , and use Eq. 2-16: 2 22 01 2 (5.08 m/s) 2 2(1.22 m/s ) vv a x x =+ Δ ¡ Δ= 10.59 m 10.6 m. =≈ (b) Using Eq. 2-11, we have 0 1 2 1 5.08 m/s 4.17 s. 1.22 m/s t a == = The deceleration time
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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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