25. The constant acceleration stated in the problem permits the use of the equations in Table 2-1. (a) We solve v= v0+ atfor the time: tvva=−=×=×01108630 109831 10(...m/s)m/ss2which is equivalent to 1.2 months.
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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.