horizontal motion formula: d = rt - webmatetcanright.macmate.me/algebra...
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d = rtHorizontal motion formula:
distance = rate • time
d = rt
d = rt – 5t2Vertical motion formula:
Horizontal motion formula:
distance = rate • time
d = rt
d = rt – 5t2Vertical motion formula:
d = distance in meters above or below starting point
Horizontal motion formula:
distance = rate • time
d = rt
d = rt – 5t2Vertical motion formula:
r = initial upward velocity in meters per second
d = distance in meters above or below starting point
Horizontal motion formula:
distance = rate • time
d = rt
d = rt – 5t2Vertical motion formula:
r = initial upward velocity in meters per second
d = distance in meters above or below starting point
t = time in seconds since object was released
Horizontal motion formula:
distance = rate • time
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
d = rt – 5t2
A person standing on the ground throws a ball upward with an initial velocity of 40 m/sec.
d = rt – 5t2
12345678
sec. calculation distance
A person standing on the ground throws a ball upward with an initial velocity of 40 m/sec.
d = rt – 5t2
1 40•1–5•1=40–5 35 m
2345678
sec. calculation distance
A person standing on the ground throws a ball upward with an initial velocity of 40 m/sec.
d = rt – 5t2
1 40•1–5•1=40–5 35 m
2 40•2–5•4=80–20 60 m
345678
sec. calculation distance
A person standing on the ground throws a ball upward with an initial velocity of 40 m/sec.
d = rt – 5t2
1 40•1–5•1=40–5 35 m
2 40•2–5•4=80–20 60 m
3 40•3–5•9=120–45 75 m
45678
sec. calculation distance
A person standing on the ground throws a ball upward with an initial velocity of 40 m/sec.
d = rt – 5t2
1 40•1–5•1=40–5 35 m
2 40•2–5•4=80–20 60 m
3 40•3–5•9=120–45 75 m
4 40•4–5•16=160–80 80 m
5678
sec. calculation distance
A person standing on the ground throws a ball upward with an initial velocity of 40 m/sec.
d = rt – 5t2
1 40•1–5•1=40–5 35 m
2 40•2–5•4=80–20 60 m
3 40•3–5•9=120–45 75 m
4 40•4–5•16=160–80 80 m
5 40•5–5•25=200–125 75 m
678
sec. calculation distance
A person standing on the ground throws a ball upward with an initial velocity of 40 m/sec.
d = rt – 5t2
1 40•1–5•1=40–5 35 m
2 40•2–5•4=80–20 60 m
3 40•3–5•9=120–45 75 m
4 40•4–5•16=160–80 80 m
5 40•5–5•25=200–125 75 m
6 40•6–5•36=240–180 60 m
78
sec. calculation distance
A person standing on the ground throws a ball upward with an initial velocity of 40 m/sec.
d = rt – 5t2
1 40•1–5•1=40–5 35 m
2 40•2–5•4=80–20 60 m
3 40•3–5•9=120–45 75 m
4 40•4–5•16=160–80 80 m
5 40•5–5•25=200–125 75 m
6 40•6–5•36=240–180 60 m
7 40•7–5•49=280–245 35 m
8
sec. calculation distance
A person standing on the ground throws a ball upward with an initial velocity of 40 m/sec.
d = rt – 5t2
1 40•1–5•1=40–5 35 m
2 40•2–5•4=80–20 60 m
3 40•3–5•9=120–45 75 m
4 40•4–5•16=160–80 80 m
5 40•5–5•25=200–125 75 m
6 40•6–5•36=240–180 60 m
7 40•7–5•49=280–245 35 m
8 40•8–5•64=320–320 0 m
sec. calculation distance
35 m1 sec
35 m1 sec
60 m2 sec
35 m1 sec
60 m2 sec
75 m3 sec
35 m1 sec
60 m2 sec
75 m3 sec
80 m in 4 seconds
35 m1 sec
60 m2 sec
75 m3 sec
80 m in 4 seconds
75 m5 sec
35 m1 sec
60 m2 sec
75 m3 sec
80 m in 4 seconds
75 m5 sec
60 m6 sec
35 m1 sec
60 m2 sec
75 m3 sec
80 m in 4 seconds
75 m5 sec
60 m6 sec
35 m7 sec
35 m1 sec
60 m2 sec
75 m3 sec
80 m in 4 seconds
75 m5 sec
60 m6 sec
35 m7 sec
on the ground in 8 sec
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit the ground?
d = rt – 5t2
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit the ground?
d = rt – 5t2
d = rt – 5t2
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit the ground?
d = rt – 5t2
d = rt – 5t2
0 = 40t – 5t2
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit the ground?
d = rt – 5t2
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit the ground?
d = rt – 5t2
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
t = 0 or 40 – 5t = 0
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit the ground?
d = rt – 5t2
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
t = 0 or 40 – 5t = 0
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit the ground?
d = rt – 5t2
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
t = 0 or 40 – 5t = 040 = 5t
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit the ground?
d = rt – 5t2
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
t = 0 or 40 – 5t = 040 = 5t8 = t
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How long will it take the ball to hit the ground?
d = rt – 5t2
d = rt – 5t2
0 = 40t – 5t2
0 = t(40 – 5t)
t = 0 or 40 – 5t = 040 = 5t8 = t
It will take 8 sec.
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5 seconds?
d = rt – 5t2
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5 seconds?
d = rt – 5t2
d = rt – 5t2
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5 seconds?
d = rt – 5t2
d = rt – 5t2
d = 40(6.5) – 5(6.5)2
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5 seconds?
d = rt – 5t2
d = rt – 5t2
d = 40(6.5) – 5(6.5)2
d = 260 – 5(42.25)
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5 seconds?
d = rt – 5t2
d = rt – 5t2
d = 40(6.5) – 5(6.5)2
d = 260 – 5(42.25)d = 260 – 211.25
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How high will the ball be after 6.5 seconds?
d = rt – 5t2
d = rt – 5t2
d = 40(6.5) – 5(6.5)2
d = 260 – 5(42.25)d = 260 – 211.25
d = 48.75 m
35 m1 sec
60 m2 sec
75 m3 sec
80 m in 4 seconds
75 m5 sec
60 m6 sec
35 m7 sec
on the ground in 8 sec
48.75 m 6.5 sec
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for the ball to be 100 m off the ground?
d = rt – 5t2
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for the ball to be 100 m off the ground?
d = rt – 5t2
d = rt – 5t2
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for the ball to be 100 m off the ground?
d = rt – 5t2
d = rt – 5t2
100 = 40t – 5t2
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for the ball to be 100 m off the ground?
d = rt – 5t2
d = rt – 5t2
100 = 40t – 5t2
5t2 – 40t + 100 = 0
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for the ball to be 100 m off the ground?
d = rt – 5t2
d = rt – 5t2
100 = 40t – 5t2
5t2 – 40t + 100 = 0 5(t2 – 8t + 20) = 0
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for the ball to be 100 m off the ground?
d = rt – 5t2
d = rt – 5t2
100 = 40t – 5t2
5t2 – 40t + 100 = 0 5(t2 – 8t + 20) = 0
t2 – 8t + 20 = 0
A person standing on the ground throws a ball upward with an initial
velocity of 40 m/sec.
How many seconds will it take for the ball to be 100 m off the ground?
d = rt – 5t2
d = rt – 5t2
100 = 40t – 5t2
5t2 – 40t + 100 = 0 5(t2 – 8t + 20) = 0
t2 – 8t + 20 = 0
The discriminant is negative so there is no answer. In this case that means that the ball never reaches
100 m off the ground.
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?
0-meters
0-meters
-10-meters
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
-10 = 1.9r – 5 • 3.61
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
-10 = 1.9r – 5 • 3.61 -10 = 1.9r – 18.05
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
-10 = 1.9r – 5 • 3.61 -10 = 1.9r – 18.05
8.05 = 1.9r
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
-10 = 1.9r – 5 • 3.61 -10 = 1.9r – 18.05
8.05 = 1.9r4.24 r
d = rt – 5t2
-10 = r • 1.9 – 5(1.9)2
-10 = 1.9r – 5 • 3.61 -10 = 1.9r – 18.05
8.05 = 1.9r4.24 r
Her initial upward velocity was about 4.24 m/sec.
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
0 = t(4.24 – 5t)
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
0 = t(4.24 – 5t) 0 = t 0 = 4.24 – 5t
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
0 = t(4.24 – 5t) 0 = t 0 = 4.24 – 5t
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
0 = t(4.24 – 5t) 0 = t 0 = 4.24 – 5t
5t = 4.24
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?How long long will it take for the diver to be
even with the platform on the way down?
0 = 4.24t – 5t2
0 = t(4.24 – 5t) 0 = t 0 = 4.24 – 5t
5t = 4.24 t = 0.848 sec.
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?What is the highest point the diver will get from
the surface of the water?
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?What is the highest point the diver will get from
the surface of the water?
0-meters
-10-meters
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?What is the highest point the diver will get from
the surface of the water?
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?What is the highest point the diver will get from
the surface of the water?
0.848 ÷ 2 = 0.424 sec.
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?What is the highest point the diver will get from
the surface of the water?
0.848 ÷ 2 = 0.424 sec. d = 4.24(0.424) – 5(0.424)2
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?What is the highest point the diver will get from
the surface of the water?
0.848 ÷ 2 = 0.424 sec. d = 4.24(0.424) – 5(0.424)2
d = 1.798 – 5(0.180)
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?What is the highest point the diver will get from
the surface of the water?
0.848 ÷ 2 = 0.424 sec. d = 4.24(0.424) – 5(0.424)2
d = 1.798 – 5(0.180) d = 1.798 – 0.899
A diver dives from a 10-meter platform by jumping up before coming back down to the
water. If the diver took 1.9 seconds to get into the water, what was her initial upward velocity?What is the highest point the diver will get from
the surface of the water?
0.848 ÷ 2 = 0.424 sec. d = 4.24(0.424) – 5(0.424)2
d = 1.798 – 5(0.180) d = 1.798 – 0.899 d = 0.899 m or about 0.9 m
The diver gets about 0.9 meters above the platform
so she is 10.9 meters above the water.
The formula, d = rt – 5t2, is often used when working with vertical
motion. The rate (r) is m/sec., so the distance (d) is in meters and the time
(t) is in seconds. The formula is a quadratic equation. A related formula
is d = rt – 16t2, where the rate is given in ft/sec.