31. We assume the periods of acceleration (duration t1) and deceleration (duration t2) are periods of constant aso that Table 2-1 can be used. Taking the direction of motion to be +xthen a1= +1.22 m/s2and a2= –1.22 m/s2. We use SI units so the velocity at t= t1is v= 305/60 = 5.08 m/s. (a) We denote Δxas the distance moved during t1, and use Eq. 2-16: 222012(5.08 m/s)2 2(1.22 m/s )vvaxx=+Δ¡Δ=10.59 m 10.6 m.=≈(b) Using Eq. 2-11, we have 01215.08 m/s4.17 s.1.22 m/sta−===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.