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review1(lastsemester)

# review1(lastsemester) - PHYSICS 231 Review problems for...

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PHY 231 1 PHYSICS 231 Review problems for midterm 1

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PHY 231 2 v(t)=v(0)+at x(t)=x(0)+v(0)t+0.5at 2 Cut problem up in 1s pieces After 1 s: v(1)=0+0x1=0 x(1)=0+0x1+0.5x0X1 2 =0 After 2 s: v(2)=v(1)+at=0+3x1=3 x(2)=x(1)+v(1)t+0.5at 2 =0+0x1+0.5x3x1 2 =1.5 After 3 s: v(3)=v(2)+at=3+2x1=5 x(3)=x(2)+v(2)t+0.5at 2 =1.5+3x1+0.5x2x1 2 =5.5 After 4 s: v(4)=v(3)+at=5-2x1=3 x(4)=x(3)+v(3)t+0.5at 2 =5.5+5x1+0.5x(-2)x1 2 =9.5 What is the displacement at t=4 s. Velocity (m/s) 3 3 1 1 1.5 By drawing: Derive v(t) diagram from a(t) diagram: red line x(t) is area under v(t) diagram:
PHY 231 3 Cross fast 3.30 m/s 6.50 m/s To cross fast: use picture b) V=6.5 m/s so t=x/v=40.3 s (but lands downstream) b) 3.30 m/s 3.30 m/s 6.50 m/s v? Velocity of water Velocity of boat needed to cancel motion of water Total velocity of the boat (I.e. available in still water) Velocity ‘left over’ for crossing river v 2 +3.30 2 =6.50 2 so v= (6.50 2 -3.30 2 )=5.6 m/s Time to cross river: t=x/v=262.0/5.6=46.8 s (use v=x/t) Cross straight a) To cross straight: use picture a)

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PHY 231 4 a) Acceleration in vertical direction is ALWAYS g=–9.81 m/s 2 ; TRUE b) See a) FALSE c) It is positive going up, negative going down: FALSE d) It is zero at the start and end and positive everywhere else: TRUE +
PHY 231 5 v x (0)=29.0 m/s 2.19 m x(t)=x 0 +v 0x t = 0+29t=29t v x (t)=v 0x = 29 y(t)=y 0 +v 0y t-0.5gt 2 = 2.19+0t-0.5x9.8t 2 =2.19-4.9t 2 v y (t)=v 0y -gt = 0-9.8t=-9.8t

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