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Unit 2 Class Notes
Honors Physics
The Kinematics Equations (1D Equations of Motion)
Day 6
Mixed Review (part II)
x = 2,224.5m ttotal = m28.72 sec
Clearly this is a free-fall problem
Remember the free-fall assumptions:V1 = 0, a = -9.8m/s2, down is negative
Solve the appropriate equations:
x 12 at
2 v1t
50 12 ( 9.8)t 2 0t
t 3.19sec
v22 v1
2 2ax
v22 02 2( 9.8)( 50)
v2 31.2 ms
What must be neglected??? AIR RESISTANCE!!!
Again, this is a free-fall problem
Remember the free-fall assumptions:V1 = 0, a = -9.8m/s2, down is negative
Solve the appropriate equations:
x 12 at
2 v1t
x0 3 12 ( 9.8)(3)2 0(3)
x0 3 44.1m
How far will it fall from 3-4 seconds?
x 12 at
2 v1t
x0 4 12 ( 9.8)(4)2 0(4)
x0 4 78.4m
d3 4 d0 4 d0 3 78.14 44.1 34.3m
During the 20th second?
d19 20 d0 20 d0 19
1960 1768.9 191.1m
Clearly this is a throw up problem
Remember the assumptions: V2 = 0 (@ top), a = -9.8m/s2, down is negative
Make sure to draw a picture and label the points appropriately.
x 12 at
2 v1t
50 12 ( 9.8)t 2 0t
t 3.19sec
What must be neglected??? AIR RESISTANCE!!! 1
2
3
4
Clearly this is a throw up problem
Remember the assumptions: V2 = 0 (@ top), a = -9.8m/s2, down is negative
Make sure to draw a picture and label the points appropriately.Choose two points, and work between these two points.
1
2
3
4
choose points 1 & 2choose points 1 & 2
choose points 1 & 4choose points 1 & 3
1
2
3
4
Points 1-2
v1 = 10 m/s
a = -9.8 m/s2
v2 = 0 m/s
02 102 2( 9.8)x
v22 v1
2 2ax
x 5.1m
v2 v1 at
0 10 ( 9.8)t
t 1.02sec
Points 1-4
a = -9.8 m/s2
x = 0 m
v1 = 10 m/s
x 12 at
2 v1t
0 12 ( 9.8)t 2 10t
t 2.04 sec
OR ….Simply double the time from point 1 to point 2.
1
2
3
4
v1 = 10 m/s
a = -9.8 m/s2
x = 5m
x 12 at
2 v1t
5 12 ( 9.8)t 2 10t
t 0.876sec, 1.165sec
Solve this quadratic equation for the times Graph y = -4.9x2 + 10x - 5
Points 1-3
0 12 ( 9.8)t 2 10t 5
Hit “2nd Trace”, “ZERO”
Left Bound, Right Bound, Guess ….It is 5 meters above the ground on the way UP and on the way DOWN.
Points 1-2
v1 40.92 fts
1
2
3
Remember your assumptions, draw a picture, pick your points.
x = x2 – x1 = 26 ft
v1 = ?
a = -32.2 ft/s2
x 1= 4 ft
x2 = 30 ft
v2 = 0
v22 v1
2 2ax
02 v12 2( 32.2)(26)
x = x2 – x1 = -4 ft
a = -32.2 ft/s2
x 1= 4 ft
x2 = 0 ft
v1 = 40.92 ft/s
x 12 at
2 v1t
4 12 ( 32.2)t 2 (40.92)t
16.1t 2 40.92t 4 0
t .094 sec, 2.64 sec
Points 1-3
This is a chase problem!!!
x2 12 at
2 v1t x1
x2N = x2O
Use the chase equation:
But it’s also a CHALLENGING chase problem. Why? Because the “new” car undergoes two different motions (speeding up and then coasting).
12 at
2 v1taccelerating
1 2 4 3 4 12 at
2 v1tcoasting
1 2 4 3 4 x1
N
12 at
2 v1t x1 O
Notice how each different motion needs to be accounted for when writing the equation.
Plug in what you know
12 at
2 v1taccelerating
1 2 4 3 4 12 at
2 v1tcoasting
1 2 4 3 4 x1
N
12 at
2 v1t x1 O
12 ataccel
2 0taccelaccelerating
1 2 4 4 3 4 4 12 (0)tcoast
2 21tcoastcoasting
1 2 4 4 4 3 4 4 4 0 12 (0)t 2 13t 0
v1 = 0
v2 = 21 m/s
t = 40 sec
v2 v1 atHow do we find “a”?
21 0 a(40)
a .525 ms2
How do we plug in for the times?
tO = t (the total time)
taccel = 40 sec (this was given)
tcoast = t – 40 (the total time minus the accel time)
Solve
12 (.525)(40)2 0(40) 1
2 (0)(t 40)2 21(t 40) 0 12 (0)t 2 13t 0
4200 0 21t 8400 0 13t 0
420 8t
t 52.5sec
13
21
v (m/s)
Time (sec)40 t
When the areas under the curves are equal, the new car has caught the older one.
Atriangle Arectangle Arectangle
12 (40)(21) (t 40)(21) 13t
t 52.5sec
THE GRAPHICAL APPROACH!!!
TONIGHTS HWNONE ENJOY!!!
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