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PHY 231
1
PHYSICS 231
Review problems for midterm 1
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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=52x1=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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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
t0.5gt
2
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This note was uploaded on 02/24/2009 for the course HNF 150 taught by Professor Thurston during the Spring '07 term at Michigan State University.
 Spring '07
 Thurston

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