46 Motion in One Dimension*P2.60Average speed of every point on the train as the first car passes Liz:∆∆xt==860573..m1.50 sms.The train has this as its instantaneous speed halfway through the 1.50 s time. Similarly, halfwaythrough the next 1.10 s, the speed of the train is 782..m1.10 s=. The time required for the speedto change from 5.73 m/s to 7.82 m/s is1215012110130...s ss()+=so the acceleration is: avtxx−=∆∆160..s2.P2.61The rate of hair growth is a velocity and the rate of its increase is an acceleration. Thenvxi=104.mm d and ax=FHGIKJ0132.mm dw. The increase in the length of the hair (i.e., displacement)during a time of t500350w dis∆∆xvt atxxix=+⋅12120 13235 05 002...mm ddmm d wdwbgafafor ∆x=48 0. mm.P2.62Let point 0 be at ground level and point 1 be at the end of the engine burn. Letpoint 2 be the highest point the rocket reaches and point 3 be just before
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Velocity, 1.60 M, instantaneous speed halfway, 1.73 km, 5.00 w, 5.73 m/s