Outline  One Dimensional Motion


 One Dimensional Motion
 Displacement is the change in position of an object. It
is represented by a directed line segment s. For motion
in the x or ydirection, it is represented by x or y, respectively.
 Speed is the distance traversed per unit time. The velocity
v of an object is a specification of both its speed
and the direction in which it is moving.
 The average velocity
of an object during a time interval t is the distance
covered during this time interval, divided by the time t.
 Instantaneous velocity is the velocity of an object at
any instant of time.
 For a case of constant velocity, the instantaneous velocity
is equal to the constant velocity at all times.
 Acceleration a is the rate at which an object's velocity
changes per unit time.
 The change in velocity may be in magnitude, or direction,
or both.
 For constant acceleration, the direction of the velocity
does not change and the rate of change of the velocity is
constant.
 Motion in OneDimension
 Relations Among Descriptive Quantities
 Velocity and Displacement
 The slope of x versus t at any instant equals the
velocity at that instant
 The area under the v versus t curve from t = 0 to
t = t equals the distance traveled, (x  x_{o})
 Velocity and Acceleration
 The slope of v versus t at any instant equals the
acceleration at that instant.
 The area under the a versus t curve from t = 0 to
t = t equals the change in velocity (v  v_{o}).
 For Constant Velocity v
where x = position at time t and x_{o}
= position at t = 0. Rearranging, x function of t, x(t)
= vt + x_{o} (2c) compare with
y function of x, y(x) = mx + b and see that a plot of
x as a function of t gives a straight line with the slope
equal to the velocity and intercept on the Xaxis equal
to x_{o}.
 Example in Fig. 1 below with v = 2.0 m/s
 For graph 1 in Fig. 1a,
x_{o} = 1.5 m and
x(t) = (2.0 m/s)t + 1.5 m
 For graph 2 in Fig. 1a,
x_{o} = 0 and x(t)
= (2.0 m/s)t
 Velocity v as a function of time t in (Fig. 1b
above). For all times,
v = 2.0 m/s. A plot of v as a function of times gives
a straight line parallel to the taxis. The area under
the v versus t curve = vt. From t = 0 to t = 1.0 s,
vt = (2.0 m/s)(1.0 s) = 2.0 m = the distance moved.
In Fig. 1a for graph 1, x_{o}
= 1.5 m and x(t = 1 s) = x_{o}
+ area under v vs t = 1.5 m + 2.0 m = 3.5 m and x
= 3.5 m at t = 1.0 s. In Fig. 1a for graph 2, x_{o}
= 0 and x = area under v vs t = 2.0 m and x = 2.0
m at t = 1.0 s. For all cases of constant velocity,
a plot of v as a function of time yields a straight
line parallel to the taxis. The slope of this line
is zero.
 In Fig. 1 c above, the acceleration a is zero for
all times. The area under a versus t is zero which
corresponds to zero change in velocity. For all cases
of constant velocity, a plot of a as a function of
time yields a straightline trough the taxis.
 For constant acceleration,
 a = (v  v_{o}/(t  0) = a constant.
(Equation
1')
Since the slope of v as a function of time t equals the
acceleration, a plot of v versus t yields a straight line
as shown in Fig. 2 above. The area under the curve of
v versus t from t = 0 to t = t is the distance moved,
or:
x  x_{o}.
From Fig. 2,
x  x_{o} = v_{o}t
+ 1/2 (v  v_{o})t (Equation
2')
Rearranging Eq. 1:
(v  v_{o}) = at (Equation
1")
Substituting Eq. 1" into Eq. 2':
x  x_{o} = v_{o}t
+ 1/2 at^{2 } (Equation
2")
If you solve for t in Eq. 1' and substitute it into Eq.
2", you find
v^{2 } = v_{o}^{2
} + 2a(x  x_{o}) (Equation
3)
 Important equations for motion with constant acceleration
 v(t) = v_{o} + at (Equation
1)
Eq.1 gives velocity v at any instant as a function
of time t.
 x(t) = x_{o} + v_{o}t
+ 1/2 at^{2} (Equation
2)
Eq. 2 gives the displacement x at any instant as a
function of time t.
 v^{2 }(x) = vo^{2
} + 2a(x  x_{o}) (Equation
3)
Eq. 3 gives the velocity v at any instant as a function
of position x.
 Example in Fig. 3 above with a = 1.0 m/s^{2
} and initial conditions, x_{o}
= 0 and v_{o} = 1.0 m/s. For this
case,
 v(t) = 1.0 m/s + 1.0 m/s^{2 }
t
 x(t) = 0 + 1.0 m/s t + 1/2(1.0 m/s^{2
}) t^{2 }
 v^{2 }(x) = (1.0 m/s)^{2
} + 2(1.0 m/s^{2 })(x
 0)
 acceleration a = slope of v vs t = 1 m/s^{2
} (Fig. 3b)
 area under v versus t graph from t = 0 to t = 1,
x(1s)  (0 s) = (0.5 +1.0)m = 1.5 m. (Fig. 3b) x(1s)
= 1.5 m + x(0 s) =
1.5 m as in Fig. 3a
 area under a versus t graph (Fig. 3c) from t = 0
to t = 1 s, v(1s)  v(0 s) = 1.0 m/s. v(1s) = 1.0
m/s + v_{o} = (1.0 + 1.0)m/s
= 2 m/s (Fig. 3b) Slope of x vs t at t = 0 equals
v_{o} = 1.0 s. Slope of x vs
t is positive and increasing (Fig. 3a) as is the velocity
(Fig 3b).
 Sample Problem. A person tosses a ball up into the
air from y_{o} = 1.0 m with an
initial velocity v_{o} = 6.0 m/s
(Fig. 4 above). Find the height to which the ball rises.
Take g = 10 m/s^{2 }.
 Approach 1. We do not know the time the ball takes
to get to its highest point or the value of y at that
point. We do know that at the highest point the ball
momentarily stops and then returns to the ground.
At the highest point, v = 0.
 For a onedimensional problem, we can indicate
directions by a "+" or "" sign.
Choose up to be positive so a = 10 m/s^{2}.
v(t) = v_{o} + at. At the
top, v= 0 and 0 = 6.0 m/s  (10 m/s2)t and t =
0.6 s.
 y(t) = y_{o} + v_{o}t
+ 1/2 at^{2 }.
y(0.6 s) =
1.0 m + 6.0 m/s(0.6 s)  1/2(10 m/s^{2})(0.6
s)^{2}= 2.8 m
 Approach 2. Use v^{2 }(y)
= v_{o}^{2
} + 2a(y  y_{o})
0 = (6 m/s)^{2 } + 2(10 m/s^{2
})(y  1.0m) or (20 m/s^{2
})(y  1.0m)
=36 m^{2 }/s^{2
}
 Note in Fig. 4a, I show the path of the ball as
it rises and falls. In Fig. 4b, I plot y as a function
of time t. Note Fig. 4b is NOT a path.
 Combination of Motions (Fig. 5a below)
 From t = 0 to t = 1, the plot of x as a function of
time is not a straight line so this is not a case of constant
velocity. The slope is positive and increasing so the
velocity is positive and increasing.
 From t = 1 to t = 2 s, the plot of x as a function of
time is a straight line. This is a case of constant positive
velocity.
 From t = 2 to t = 3 s,the plot of x as a function of
time is not a straight line so this is not a case of constant
velocity. The slope is positive, but decreasing so the
velocity is positive and decreasing.
 At t = 3 s, the slope of x versus t is zero. The velocity
at t = 3 s is zero. From t = 3s to t = 4 s, the slope
is negative and increasing. The velocity is in the negative
Xdirection and it is increasing.
 Fig. 5b below is a plot of velocity v as a function
of time t. The analysis given in 4 above agrees with the
results of this plot. We find the values of x for any
time t from the area under the v versus t curve. As shown
in Fig. 5b, the area under v vs. t from t = 0 to t = 1
s is 1.5 m and x(1s) = 1.5 m. The area under v vs. t from
t = 0 to t = 2 s is 3.5 m and x(2s) = 3.5 m. The area
under v vs. t from t = 0 to t = 3 s is 4.5 m and x(3s)
= 4.5 m. The area under v vs. t from t = 0 to t = 4 s
is (1.5 + 2.0 + 1.0  1.0)m = 3.5 m and x(4s) = 3.5 m.
 The slopes of v versus t for the various time intervals
are given in Fig. 5b.
From t = 0 to t = 1.0 s,
dv/dt = a = 1 m/s^{2}
From t = 1 to t = 2.0 s,
dv/dt = a = 0
From t = 2 to t = 4.0 s,
dv/dt = a = 2.0 m/s^{2}
 The areas under a versus t gives the change in velocity.
The initial velocity was 1.0 m/s. The change in velocity
from t = 0 to t = 1.0 s is 1 m/s. v(1s) = 1 m/s + 1 m/s
= 2 m/s. The change in velocity from t = 0 to t = 2.0
s is (1 + 0) m/s = 1 m/s v(2s) = 1 m/s + 1 m/s = 2 m/s.
The change in velocity from t = 0 to t = 3.0 s is (1 +
0  2.0)m/s = 1 m/s. v(3s) = 1 m/s  1 m/s = 0 m/s. The
change in velocity from t = 0 to t = 4.0 s is (1 + 0 
2.0  2.0)m/s = 3.0 m/s v(3s) = 1 m/s  3 m/s = 2 m/s.
 Motion with Constant Velocity as a Special Case of Constant
Acceleration
 For constant acceleration,
 v(t) = v_{o} + at (1)
 x(t) = x_{o} + v_{o}t
+ 1/2 at^{2 } (2)
 v^{2 }(x) = v_{o}^{2
}_{o}) (3)
 For constant velocity a = 0 and the equations above
become
 v(t) = v_{o} (1c)
 x(t) = x_{o} + v_{o}t
(2c)
 v^{2 }(x) = v_{o}^{2
} (3c) Notice for v(t) = v_{o}
= constant, Eq. 2c above is equivalent to Eq. 2c in
Section B1 on page 2.
 Units
 Dimensional Analysis (Units are your friends!)
 Given an object moving with constant acceleration a
= 4.0 m/s^{2 }, an initial velocity
of 5.0 m/s, and an initial position at t = 0 of 1.0 m,
find its position at t = 2.0 s.
 Let's say you make a mistake and write x(t) as
 x(t) =? x_{o} + v_{o}t
+ 1/2 at
 Substitute in the values that are given with their
units
 x(2.0 s) =? 1.0 m + (5.0 m/s)(2.0 s) + 1/2(4.0
m/s^{2 })(2.0 s)
 This gives x(2.0 s) ‚ 1.0 m + 10 m + 8 m/s which
cannot be true because m ≠ m/s
 Conversion of Units
 Change 60 miles/hour to ft/s
 1 mile = 5280 ft or 5280 ft/1 mile = 1
 1 hour = 3600 s or 1 hour/3600 s = 1
 You do not change a quantity if you multiply it by 1
 Thus 60 mile/hour x 5280 ft/mile x 1 hour/3600 s = 60(5280/3600)
ft/s =
88 ft/s. Notice I have arranged the quantities equal to 1
so that the units of miles and hour cancel
 Practice Problems in 104
Problem Set for One Dimensional Motion:
16, 10,
1216, and 19.

