30_Physics ProblemsTechnical Physics

30_Physics ProblemsTechnical Physics - Chapter 2 (d) 1 1 a1...

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Chapter 2 31 (d) (i) xa t t 11 22 0 1 2 1 2 33 =+ = . ms 2 ch or xt 1 2 167 = . m s 2 (ii) 2 1 2 15 50 0 50 15 =( ) + () s m s m s s af or 2 50 375 = ms m a f (iii) For 40 50 s ≤≤ t , x vt t at t 2 0 1 2 40 50 40 = = F H G I K J + + area under vs from to 40 s m s s a f or t 3 2 375 1 250 1 2 5 0 40 50 40 + −+ m m m s s m s s 2 . ej a f bg a f which reduces to t 3 2 250 2 5 4 375 =− m 2 .. (e) v == = total displacement total elapsed time m s 1875 50 37 5 . P2.25 (a) Compare the position equation t 200 300 400 2 . . . to the general form xx v t a t fii =+ + 1 2 2 to recognize that x i = 200 m , v i = 300 . m s, and a = 800 m s 2 . The velocity equation, vv a t fi , is then f = 2 . The particle changes direction when v f = 0, which occurs at t = 3 8 s . The position at this time is: x F H G I K J F H G I K J = 3 8 3 8 256 2 . . m m s
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