Uniform Acceleration

1

• Uniform Acceleration
• Earlier we learned that acceleration was the rate
  at which ____________ changed with respect to
  time. Therefore, if either the speed or
  direction of an object changed during motion,
  then there would be an __________________.
  Example If an object has an initial velocity of
  0 m/s and accelerates in a straight line at a
  rate of 2 m/s2, we could see that after 1 second
  the magnitude of the objects velocity would be
  ____ m/s, after 2 s, ____ m/s, after 3 s, _____
  m/s and after 4.0 s, ____m/s.
• See the figure below.
• 0s 1s 2s
  3s 4s
• 0 m/s 2 m/s 4 m/s
  6 m/s 8 m/s
• Notice that for an object with a constant
  acceleration, the displacement of the object
  between equal time intervals traveled
  _______________________.
• Solving Problems for Moving Objects with Uniform
  Acceleration
• Example 1 Lets say that an object has an
  initial velocity of 10.0 m/s and accelerates for
  3.0 seconds. The objects velocity at the end of
  those 3.0 seconds is 40.0 m/s. Here the
  acceleration and displacement is not known. Lets
  see if we can determine the displacement traveled
  during the 3.0 second time interval using known
  equations for average velocity.
• Step 1 We should be able to see that the
  objects average velocity during those 3.0s was
  _______ m/s. _
• v (vf vo) / 2

velocity
acceleration
4
6
2
8
increases
25.0
2

• Step 2 The displacement could than be found
  because we know that the object is in essence
  traveling _______ m/s for 3.0 s. This would be
  ____________ m.
• _
• 
  v d/t
• d v t
• d 25.000 m/s(3.0 s)
• d 75.0 m
• We could have solved this problem in 1 step by
  combining the 2 equations used above to obtain a
  new equation.
• v (vf vo ) d / t
• 2
• d (vf vo) t
• 2
• d (40.0 m/s 10.0 m/s) (3.0 s)
• 2
• d (50.0 m/s) (3.0 s) / 2
• d 75 m

25.0
75
_

_

3

• Example 2 Lets say that an object has an
  initial velocity of 10.0 m/s and accelerates at a
  rate of
• 5.00 m/s2 for 3.00 s, determine the displacement
  of the object using known equations for velocity
  and acceleration. Here the final velocity and
  displacement are not known.
• Step 1 Here we can see that the final velocity
  would be ________ m/s.
• a (vf vo)
• t
• vf at vo
• vf (5.00 m/s2)(3.00 s) 10.0 m/s
• vf (15.000 m/s) 10.0 m/s
• vf 25.0 m/s
• Step 2 The average velocity of the object would
  then be ________ m/s.
• v (vf vo)
• 2
• v (25.000 m/s 10.0 m/s) / 2
• v (35.000 m/s) / 2
• v 17.5 m/s

25.0


17.5
_
_
_
_

4

• Step 3 The displacement of the object would
  then be ________ m
• v d/t
• d v t
• d (17.500 m/s)(3.00 s)
• d 52.5 m
• a 5.00 m/s2
• t 3.00 s
• vo 10.0 m/s
• vf wasnt given
• d we were to determine

52.5
_
_

Instead of using this 3 step thought process,
this problem can be calculated using 1 equation.
The equation can be derived using the equations
shown above.
vf vo d 2 t
a vf - vo t
vf at vo
at vo vo d 2 t
d ½ at2 vot
d ½ (5.00 m/s2)(3.00s)2 (10.0 m/s)(3.00s)
d 22.500 m 30.000 m
d 52.5 m
5

• Example 3 What if an object has an initial
  velocity of 10.0 m/s and a final velocity of 60.0
  m/s at an acceleration of 10.0 m/s2. Determine
  the displacement of the object using known
  equations for average velocity and acceleration.
  Here, the time isnt known.
• Step 1 Here, we could see that the average
  velocity would be _____ m/s.
• _

35.0
_
v (vf vo)
2
_
v (60.0 m/s 10.0 m/s)
2
_
v (70.0 m/s)
2
_
v 35.0 m/s
5.00
Step 2 We can also see that the object
accelerates for ________ s. a (vf - vo) t
t (vf - vo)
a
t (60.0 m/s 10.0 m/s)
10.0 m/s2
t (50.0 m/s)
10.0 m/s2
t 5.00 s
6

• Step 3 Therefore, the object travels __________
  m.

175
_
d v t
d (35.000 m/s) (5.0000 s)
d 175 m
This problem can be solved using 1 equation
derived by combining the equations used above.
a (vf - vo) t
vf vo d 2 t
vo 10.0 m/s
vf 60.0 m/s
d vf2 vo2 2a
t (vf - vo) a
a 10.0 m/s2
vf vo d 2 (vf-vo) a
t not given
d (60.0 m/s)2 (10.0m/s)2 2(10.0 m/s2)
d ?
(vf vo) (vf vo) ad 2
d 3600.0 m2/s2 100.0 m2 /s2 2(10.0 m/s2)
(vf vo) (vf vo) 2 ad
d (3500.0 m2/s / 2(10.0 m/s2))
vf 2 vo2 2 ad
d 175 m
7

• Acceleration and Free Fall (pg. 17-24 pg 47-51)
• In free fall an object travels up or down with
  only the force of ___________ acting upon it.
  The force of gravity will cause the object to
  accelerate towards earth at a rate of 9.80 m/s2,
  meaning that every second the objects speed
  changes by ________ m/s (that is if we can
  neglect air resistance).
• Sample problems 1 An HSE diver steps off the
  high dive and strikes the water below 2.1 seconds
  later. Determine divers speed at the time of
  impact with the water.
• vo 0 m/s
• t 2.1 s
• a 9.80 m/s2
• vf ?
• Sample Problem 2 Determine the displacement of
  the diver.
• d ½ at2 vot
• d ½ (9.80 m/s2)(2.1s)2
• d 22 m
• Sample Problem 3 King Kong holds Fay Wray at
  the top of the Empire State Building. Her shoe
  falls off and strikes the ground at 80.1 m/s.
  Determine the displacement of the shoe during the
  fall.
• a 9.80 m/s2
• vf 80.1 m/s
• vo 0 m/s
• d ?

gravity
9.80
a vf vo / t vf at vo vf (9.80 m/s2)
(2.1s) vf 21 m/s


vf2 vo2 2ad d (vf2 vo2) / 2a d (80.1
m/s)2 / (2(9.80 m/s2)) d 327 m

8

• When an object is thrown straight upward (up
  being the negative direction), the magnitude of
  the velocity ___________. Upward velocity values
  are given a __ sign (here the negative represents
  direction). Downward velocity values are given a
  __ sign. Further, upward displacements are __
  while downward displacements are ___ (throwing up
  is a negative thing!)
• Example An object is thrown straight up with
  the magnitude of its velocity being 29.4 m/s (the
  complete velocity with sign would be ________
  m/s).
• 1.00 seconds after being thrown the object will
  be traveling 9.80 m/s slower than it did the
  previous second or ________ m/s (it travels
  slower due to traveling in the opposite the
  direction of the gravitational acceleration which
  acts toward the earths center).
• 2.00 seconds after being thrown the object will
  be traveling 9.80 m/s slower than it did the
  previous second or ______ m/s.
• 3.00 seconds after being thrown the object will
  be traveling 9.80 m/s slower than it did the
  previous second or ______ m/s.
• 4.00 seconds after being thrown the object will
  be traveling 9.80 m/s faster than it did the
  previous second or _______m/s (it travels faster
  due to traveling in the downward direction or in
  the same direction as the gravitational
  acceleration).
• 5.00 seconds after being thrown the object will
  be traveling 9.80 m/s faster than it did the
  previous second or ________m/s
• 6.00 seconds after being thrown the object will
  be traveling 9.80 m/s faster than it did the
  previous second or _________m/s. (see fig. 2.6
  pg. 18 3.8 pg. 50). Here the object would
  return to your hand with the _________ magnitude
  of velocity as when it left your hand but a
  different _______ due to traveling in the
  opposite direction.
• Recall that the slope of a velocity vs. time
  graph yields a value for the _________________.
• Here, the acceleration (or slope of the line) of
  the object is seen to be 9.80 m/s2 (a positive
  value at all times). Even at the _________ of
  the objects path when the velocity is zero, the
  acceleration due to gravity is still positive
  9.80 m/s2. Therefore we will __________use the
  acceleration of gravity to be 9.80m/s2.

-
decreases

-

-29.4
-19.6
-9.8
0.0
9.8
19.6
29.4
same
sign
acceleration
top
always
9

• m (y2-y1)/(x2-x1)
• points (0.00 s, -29.4 m/s) and (3.00 s, 0.00
  m/s)
• y2 and x1 cancel so,
• m -y1 / x2
• m -(-29.4 m/s) / 3.00 s
• m 9.80 m/s2

Sample Problem 4 A Geyser in Yellowstone
National Park is capable of shooting water from
the ground with a speed of 48.0 m/s. Determine
the maximum upward displacement of the
water.
d ? vo - 48.0 m/s a 9.80 m/s2 vf
0 m/s
d (vf2 vo2) / 2a d (- 48.0 m/s)2 /
2(9.80m/s2) d - 118 m
10

• Sample Problem 5 A ball is thrown straight
  upward at 39.2 m/s. Determine the velocity and
  displacement 1.5s , 4.0s and 6.5 s after being
  thrown.
• vo -39.2 m/s vf at vo d
  ½ at2 vot
• a 9.80 m/s2 vf (9.80 m/s2)(1.5s) (-39.2
  m/s) d ½ (9.80 m/s2)(1.5s)2
  (-39.2m/s)(1.5s)
• t 1.5 s vf 14.70 m/s 39.2 m/s d
  11.025 m 58.80 m
• vf ? vf -25 m/s (-24.5 m/s) d
  -48 m
• d ?
• vo -39.2 m/s vf at vo d
  ½ at2 vot
• a 9.80 m/s2 vf (9.80 m/s2)( 4.0s) (-39.2
  m/s) d ½ (9.80 m/s2)(4.0s)2 (-39.2
  m/s)(4.0s)
• t 4.0 s vf 39.20 m/s 39.2 m/s d
  78.4 m 156.8 m
• vf ? vf 0. m/s d -80
  m
• d ?
• vo -39.2 m/s vf at vo d ½ at2
  vot
• a 9.80 m/s2 vf (9.80 m/s2)( 6.5s) (-39.2
  m/s) d ½ (9.80 m/s2)(6.5s)2 (-39.2
  m/s)(6.5s)
• t 6.5 s vf 63.70 m/s 39.2 m/s d
  207 m 254.8 m
• vf ? vf 25 m/s d -50
  m (-48 m)
• d ?