 Phyllis Fleming Physics Physics 107
 Outline - One Dimensional Motion
1. Language of Motion

1. Displacement is the change in position of an object. It is represented by a directed line segment s. For motion in the x or y-direction, it is represented by x or y, respectively.

2. 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.

1. The average velocity of an object during a time interval t is the distance covered during this time interval, divided by the time t.

2. 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.

3. Acceleration a is the rate at which an object’s velocity changes per unit time.

1. The change in velocity may be in magnitude, or direction, or both.

2. For constant acceleration, the direction of the velocity does not change and the rate of change of the velocity is constant.

2. Motion in One-Dimension

1. Relations Among Descriptive Quantities

1. Velocity and Displacement

1. dx/dt = v. The slope of x versus t at any instant equals the velocity at that instant

2. Separating variables : dx = v dt

Integrating: distance moved equals:

 x – xo = = area under v vs t from t = 0 to t = t.

2. Velocity and Acceleration

1. dv/dt = a. The slope of v versus t at any instant equals the acceleration at that instant.

2. Separating variables: dv = a dt

Integrating: change in velocity equals:

 v – vo = = area under a vs t curve from t = 0 to t = t.

2. For Constant Velocity v

1. where x = position at time t and xo = position at t = 0.

Rearranging, x function of t,
x(t) = vt + xo                 (Equation2c)
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 X-axis equal to xo.

2. Example in Fig. 1 below with v = 2.0 m/s

1. For graph 1 in Fig. 1a, xo = 1.5 m and x(t) = (2.0 m/s)t + 1.5 m

2. For graph 2 in Fig. 1a, xo = 0 and x(t) = (2.0 m/s)t

3. Velocity v as a function of time t in (Fig. 1b below). For all times, v = 2.0 m/s. A plot of v as a function of times gives a straight line parallel to the t-axis. 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, xo = 1.5 m and x(t = 1 s) = xo + 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, xo = 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 t-axis. The slope of this line is zero. 4. 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 straight-line through the t-axis.

3. For constant acceleration,

1. a = (v - vo)/(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 below. The area under the curve of v versus t from t = 0 to t = t is the distance moved x - xo. From Fig. 2 above,
x – xo = vot + 1/2(v – vo)t               (Equation 2’)
Rearranging Eq. 1':
(v – vo) = at                         (Equation 1”)
Substituting Eq. 1” into Eq. (2’):
x – xo = vot + 1/2 at2              (Equation 2”)
If you solve for t in Eq. 1’ and substitute it into Eq. 2”, you find:
v2 = vo2 + 2a(x – xo)             (Equation 3)

2. Important equations for motion with constant acceleration

1. v(t) = vo + at                                     (Equation 1)
Eq.1 gives velocity v at any instant as a function of time t

2. x(t) = xo + vot + 1/2 at2                               (Equation 2)
Eq. 2 gives the displacement x at any instant as a function of time t

3. v2(x) = vo2 + 2a(x - xo)                       (Equation 3)
Eq. 3 gives the velocity v as a function of position x

3. Example in Fig. 3 below with a = 1.0 m/s2 and initial conditions, xo = 0 and vo = 1.0 m/s.  For this case,

1. v(t) = 1.0 m/s + 1.0 m/s2 t

2. x(t) = 0 + 1.0 m/s t + 1/2(1.0 m/s2) t2

3. v2(x) = (1.0 m/s)2 + 2(1.0 m/s2)(x - 0)

4. acceleration a = slope of v vs t = 1 m/s2 (Fig. 3b)

5. area under v versus t graph from t = 0 to t = 1,
x(1s) - x(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

6. 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 + vo = (1.0 + 1.0)m/s = 2 m/s (Fig. 3b)

7. Slope of x vs t at t = 0 equals vo = 1.0 s. Slope of x vs t is positive and increasing (Fig. 3a) as is the velocity (Fig 3b). 4. Sample Problem. A person tosses a ball up into the air from yo = 1.0 m with an initial velocity vo = 6.0 m/s (Fig. 4 below). Find the height to which the ball rises. Take g = 10 m/s2. 1. 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.

1. For a one-dimensional problem, we can indicate directions by a + or - sign. Choose up to be positive so a = -10 m/s2. v(t) = vo + at. At the top, v= 0 and 0 = 6.0 m/s –(10 m/s2)t and t = 0.6 s.

2. y(t) = yo + vot + 1/2 at2
y(0.6 s) = 1.0 m + 6.0 m/s(0.6 s) - 1/2(10 m/s2)(0.6 s)2
= 2.8 m

2. Approach 2:
Use v2(y) = vo2 + 2a(y - yo) 0 = (6 m/s)2 + 2(-10 m/s2)(y - 1.0m) or (20 m/s2)(y – 1.0m) =36 m2/s2 and y - 1.0 m = 1.8 m
or y = 2.8 m.

3. 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.

4. Combination of Motions (Fig. 5a below) 1. 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.

2. 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.

3. 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.

4. 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 X-direction and it is increasing.

5. 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.  6. The slopes of v versus t for the various time intervals are given in Fig. 5b above:

From t = 0 to t = 1.0 s, dv/dt = a = 1 m/s2.
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/s2.

7. The areas under a versus t (Fig. 5c above) 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.

5. Motion with Constant Velocity as a Special Case of Constant Acceleration

1. For constant acceleration,

1. v(t) = vo + at                                      (Equation 1)

2. x(t) = xo + vot + 1/2 at2                      (Equation 2)

3. v2(x) = vo2 + 2a(x – xo)                     (Equation 3)

2. For constant velocity a = 0 and the equations above become

1. v(t) = vo                                   (Equation 1c)

2. x(t) = xo + vot                              (Equation 2c)

3. v2(x) = vo2                                 (Equation 3c)

Notice for v(t) = vo = constant, Equation 2c above is equivalent to Equation 2c in Section II, B, 1 above.

6. Sample problems in 107 Problem Set for One Dimensional Motion: 1-15.

3. Units

1. Dimensional Analysis (Units are your friends!)

1. Given an object moving with a constant acceleration a = 4.0 m/s2, 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.

2. Let's say you make a mistake and write x(t) as

1. x(t) =? xo + vot + 1/2 at

2. Substitute in the values that are given with their units

3. x(2.0 s) =? 1.0 m + (5.0 m/s)(2.0 s) + 1/2(4.0 m/s2)(2.0 s)

4. This gives x(2.0 s) ≠ 1.0 m + 10 m + 8 m/s which cannot be true because m ≠ m/s

2. Conversion of Units

1. Change 60 miles/hour to ft/s

2. 1 mile = 5280 ft or 5280 ft/1 mile = 1

3. 1 hour = 3600 s or 1 hour/3600 s = 1

4. You do not change a quantity if you multiply it by 1

5. 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.

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 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming October 8, 2002 April 15, 2003