Chapter

One-Dimesnsional Kinematics

Blinn College - Physics 1401 - Terry Honan

Kinematics is the study of motion. This chapter will introduce the basic definitions of kinematics. The definitions of the velocity and acceleration will require the introduction of the basic notions of calculus, most specifically the derivative. We will also consider in detail the simple special cases of motion with constant velocity and constant acceleration. Free fall will be discussed as an example of motion with constant acceleration.

. - The General Problem

By one dimensional motion we mean motion constrained to a line.  As examples, consider a car driving on a straight road or the vertical motion of an elevator.  The problem of motion in two or three dimensions will be discussed in the next chapter.

Position as a Function of Time

To mathematically define position we need to attach a real number line (the x-axis) to the line of motion.  To do this there are two arbitrary choices; we must choose the x=0 position and then we must choose the positive direction.

Any 1D motion can be represented graphically.  Time is the independent variable, so it will be the horizontal axis.  We will then consider graphs of x as a function of time, where x is the vertical axis.

Units:  The SI unit for distance is:  (QuantityUnit)=meter

Velocity and the Derivative

Average Velocity

If a car drives 130(QuantityUnit)  in 2 hours, we can calculate a velocity of 65 mi/hr. This is not necessarily what the speedometer would read; the speedometer reads the magnitude of the instantaneous velocity. In this case 65 mi/hr is what we call the average velocity.

We will define the average velocity by

(Subscript[v, ave])=((Δ x)/(Δ t))

where Δ (Delta) generally will represent the final value minus the initial value

Δ x=(Subscript[x, f])-(Subscript[x, i])  and   Δ t=(Subscript[t, f])-(Subscript[t, i]).

Note that the average velocity corresponds to two times (Subscript[t, i]) and (Subscript[t, f]), and (Subscript[x, i]) and (Subscript[x, f]) are the positions at the two times. In a graph of x vs. t the average velocity has the interpretation as the slope of the secant line between the two points ((Subscript[t, i]),(Subscript[x, i])) and ((Subscript[t, f]),(Subscript[x, f])).

Units:  The SI unit for velocity is: (QuantityUnit)/(QuantityUnit)

The average velocity is not the distance per unit time, it is the displacement per unit time. The distinction is important. If you move along a line from an initial position to a different position a distance d away and then back to the initial position, your displacement is zero since Δ x=(Subscript[x, f])-(Subscript[x, i])=0 but the total distance travelled is 2 d. We may define the average speed as the total distance divided by the total time.

Average Speed=((total distance)/(total time))

Instantaneous Velocity

The instantaneous velocity refers to a single time t. Take the position at t to be x. We can then consider a later time t+Δ t,  where the position is  x+Δ x. The average velocity between these two times is Δ x/Δ t. To get the instantaneous velocity we let Δ t become small; we do this by taking the limit as Δ t->0. This gives the derivative of calculus; instantaneous velocity is the time rate of change (or time derivative) of position.

v=(lim_(Δ t->0)) ((Δ x)/(Δ t))

In a calculus class, or a calculus-based physics class, such derivatives are used for calculations; in this class we will only use expressions like this to define things and never in calculations.

The graphical interpretation of the instantaneous velocity is simple. The average velocity is the slope of the secant lines. If we consider the secant lines corresponding to different Δ t values then as Δ t->0,  these secant lines approach what we call the tangent line. The velocity at t is then the slope of that tangent line. When we refer to the slope of a graph at some time, we mean the slope of the line tangent to the graph at that time.

The direction of motion in one dimension is given by the sign of the velocity. The speed is the velocity without its direction and is thus the absolute value of the velocity.

speed=|v|

Example . - Graphical Analysis of x vs. t

Solution

(Subscript[v, ave])=((Δ x)/(Δ t))=(((Subscript[x, f])-(Subscript[x, i]))/((Subscript[t, f])-(Subscript[t, i])))=((x(3(QuantityUnit))-x(1(QuantityUnit)))/(3(QuantityUnit)-0(QuantityUnit)))=((8(QuantityUnit)-2(QuantityUnit))/(3(QuantityUnit)-0(QuantityUnit)))=2(((QuantityUnit))/((QuantityUnit)))

Solution

(Subscript[v, ave])=((Δ x)/(Δ t))=(((Subscript[x, f])-(Subscript[x, i]))/((Subscript[t, f])-(Subscript[t, i])))=((x(6(QuantityUnit))-x(2(QuantityUnit)))/(6(QuantityUnit)-2(QuantityUnit)))=((2(QuantityUnit)-4(QuantityUnit))/(6(QuantityUnit)-2(QuantityUnit)))=-(1/2)(((QuantityUnit))/((QuantityUnit)))

Solution

(Subscript[v, ave])=((Δ x)/(Δ t))=(((Subscript[x, f])-(Subscript[x, i]))/((Subscript[t, f])-(Subscript[t, i])))=((x(6(QuantityUnit))-x(0(QuantityUnit)))/(6(QuantityUnit)-0(QuantityUnit)))=((2(QuantityUnit)-2(QuantityUnit))/(6(QuantityUnit)-4(QuantityUnit)))=0

Solution

Average Speed=((total distance)/(total time))

Average Speed=((6(QuantityUnit)+6(QuantityUnit))/(6(QuantityUnit)))=2(((QuantityUnit))/((QuantityUnit)))

Solution

v(1(QuantityUnit))=(Subscript[v, ave])=((x(2(QuantityUnit))-x(0(QuantityUnit)))/(2(QuantityUnit)-0(QuantityUnit)))=((4(QuantityUnit)-2(QuantityUnit))/(2(QuantityUnit)))=1(((QuantityUnit))/((QuantityUnit)))

Solution

v(4(QuantityUnit))=(Subscript[v, ave])=((x(6(QuantityUnit))-x(3(QuantityUnit)))/(6(QuantityUnit)-3(QuantityUnit)))=((2(QuantityUnit)-8(QuantityUnit))/(3(QuantityUnit)))=-2(((QuantityUnit))/((QuantityUnit)))

Solution

|v|=|v(4(QuantityUnit))|=2(((QuantityUnit))/((QuantityUnit)))

Displacement and Velocity vs. Time Graphs

In the graph of v vs. t above, let us first concentrate on the small time interval Δ t.  From the definition of velocity, we know that for a small Δ t something will move the small displacement Δ x=v Δ t. In the graph above, that distance Δ x is just the area of the rectangle between t and t+Δ t. If we want the total distance traveled between (Subscript[t, i]) and (Subscript[t, f]), it will then be the sum of all the small areas between  (Subscript[t, i]) and (Subscript[t, f]). When we talk about area this way it is implied that when the function is below the axis, meaning that the velocity is negative, the contribution to the area Δ x=v Δ t is negative.

Δ x=∑ v Δ t =Area under v vs. t  (Δ t is small)

Acceleration

Acceleration is to velocity as the velocity is to the position. Velocity is the time rate of change of position, so acceleration is the time rate of change of the velocity.

Average Acceleration

Since the average velocity is related to the position by  (Subscript[v, ave])=Δ x/Δ t  we can similarly write the average acceleration in terms of the velocity by

(Subscript[a, ave])=((Δ v)/(Δ t)).

We can think of average acceleration graphically as the slope of the secant lines of a v vs. t graph.

Units:  The SI unit for acceleration is: (QuantityUnit)/(((QuantityUnit))^2)

Example . - Graphical Analysis of x vs. t (continued)

Solution

(Subscript[a, ave])=((Δ v)/(Δ t))=(((Subscript[v, f])-(Subscript[v, i]))/((Subscript[t, f])-(Subscript[t, i])))=((v(4(QuantityUnit))-v(1(QuantityUnit)))/(4(QuantityUnit)-1(QuantityUnit)))=((-2(QuantityUnit)/(QuantityUnit)-1(QuantityUnit)/(QuantityUnit))/(3(QuantityUnit)))=-1 (((QuantityUnit))/(((QuantityUnit))^2))

Instantaneous Acceleration

The instantaneous acceleration (or just acceleration) is the time rate of change (or time derivative) of the velocity.

a=(lim_(Δ t->0)) ((Δ v)/(Δ t))

. - Constant Velocity and Acceleration

Now that we have considered the general problem of one dimensional kinematics we can now consider special cases, first constant velocity, then constant acceleration.  An important case of constant acceleration is free fall.

Constant Velocity

If velocity is a constant then the acceleration is zero, since the slope of a constant function is zero. Since velocity is the slope of the x vs. t graph, the graph must be a straight line. y=m x+b, where as usual m is the slope and b is the y-intercept. For  x vs. t this becomes: x(t)=v t+(Subscript[x, 0])where the x-intercept is the initial position (Subscript[x, 0]); this is the position at t=0, (Subscript[x, 0])=x(0). Rearranging the terms we get:

x(t)=(Subscript[x, 0])+v t

For this discussion, we are choosing the convention that the initial time is zero. Also, we will label the final time and position as just t and x: (Subscript[t, i])=0, (Subscript[t, f])=t, (Subscript[x, i])=(Subscript[x, 0])  and  (Subscript[x, f])=x. This gives:  Δ x=(Subscript[x, f])-(Subscript[x, i])=x-(Subscript[x, 0]).

x=(Subscript[x, 0])+v t  or  Δ x=v t.

This is a simple expression; for constant velocity, the distance is the product of the velocity and time.

Constant Acceleration

If the acceleration is a constant then to get the velocity we repeat the preceding procedure for going from a constant velocity to the position. Acceleration is the slope of the v vs. t graph. As before, define the initial velocity (Subscript[v, 0]) to be the velocity at t=0, (Subscript[v, 0])=v(0) and the final time and velocity become t and v. The  v vs. t graph becomes a line with slope a and v-intercept (Subscript[v, 0]):

v(t)=(Subscript[v, 0])+a t

We will now derive a set of four equations for motion with constant acceleration, where the expression above is our first.

Before finding the second expression, recall that for a graph of v vs. t, the area under the curve is the displacement, Δ x. Since the graph is a straight line the area under the curve is the area of a trapezoid. If we consider some interval from (Subscript[t, i]) to (Subscript[t, f]) then we get the following graph.

The area of a trapezoid with base b and heights (Subscript[h, i]) and (Subscript[h, f]) is the base times the average of the heights: Area=b (((Subscript[h, i])+(Subscript[h, f]))/2). In this case, it becomes: Δ x=(1/2)((Subscript[v, i])+(Subscript[v, f]))·Δ t . To understand this formula for area compare the areas of the trapezoid and the rectangle whose  height is the average of the trapezoid’s heights, as shown above. Applying (Subscript[v, ave])=Δ x/Δ t so this we see that, for motion with constant acceleration only, the average velocity at the midpoint of an interval is the average of the initial and final velocities.

(Subscript[v, ave])=((Δ x)/(Δ t))=(1/2)((Subscript[v, i])+(Subscript[v, f]))

To find the second equation for constant acceleration, let us rewrite Δ x=(1/2)((Subscript[v, i])+(Subscript[v, f]))·Δ t using the convention: (Subscript[t, i])=0, (Subscript[t, f])=t, (Subscript[v, i])=(Subscript[v, 0]) and (Subscript[v, f])=v. This gives:

Δ x=(1/2)((Subscript[v, 0])+v) t .

Using this as the second expression and v=(Subscript[v, 0])+a t as the first, we can find the third expression by eliminating v and the fourth equation by eliminating t. Eliminating v gives

Δ x=(1/2)((Subscript[v, 0])+v) t=(1/2)((Subscript[v, 0])+((Subscript[v, 0])+a t)) t=(Subscript[v, 0]) t+(1/2)a (t^2)

To write x as a function of t we use Δ x=x-(Subscript[x, 0]) to get out third equation.

x(t)=(Subscript[x, 0])+(Subscript[v, 0]) t+(1/2)a (t^2).

Note that for constant acceleration, the graph of x vs. t is a parabola. (Recall that a parabola in y vs. x has the form y=a (x^2)+b x+c.)

To get the fourth equation eliminate t from from the first two equations. Solve for t with the first equation

v=(Subscript[v, 0])+a t  ⟹  t=((v-(Subscript[v, 0]))/a)

and then plug into the second.

Δ x=(1/2)((Subscript[v, 0])+v) t=(1/2)((Subscript[v, 0])+v)((v-(Subscript[v, 0]))/a)  ⟹ 2 a Δ x=(v^2)-(Subscript[v, 0]^2)

where (A+B)(A-B)=(A^2)-(B^2) was used.

With this we have derived a set of four equations for kinematics with constant acceleration. These relate the variables t, Δ x, (Subscript[v, 0]), v and a. These will be useful for a large class of problems in this chapter.

Constant Acceleration Equations

({{v=(Subscript[v, 0])+a t}, {Δ x=(1/2)((Subscript[v, 0])+v) t}, {Δ x=(Subscript[v, 0]) t+(1/2)a (t^2)}, {(v^2)=(Subscript[v, 0]^2)+2 a Δ x}})

Example . - Decelerating Car

Solution

(Subscript[v, 0])=30.0(QuantityUnit)/(QuantityUnit) ,  v=0 , Δ x=150.(QuantityUnit)  and  a=?

(v^2)=(Subscript[v, 0]^2)+2 a Δ x   ⟹   a=-((Subscript[v, 0]^2)/(2 Δ x))=-3.00(QuantityUnit)/(((QuantityUnit))^2)

Solution

Δ x=(1/2)((Subscript[v, 0])+v) t   ⟹   t=((2 Δ x)/(Subscript[v, 0]))=10.0(QuantityUnit)

Free Fall

Free fall is one dimensional motion under the influence of only gravity.  Assuming that only gravity acts implies that we are ignoring any friction effects.  We will choose the convention that up is the positive direction.  Also, we will take y as the position variable; this will be consistent with our later usage where y is typically taken as the upward vertical variable. Galileo discovered that the acceleration of all bodies in the presence of gravity (ignoring air resistance) is the same.  The value of the downward acceleration is

g=9.80 (m/(s^2))=32.0(ft/(s^2)) .

Since up is the positive y direction and the acceleration is downward we take the acceleration to be:

a=-g

Using this value of a and replacing x with y takes the constant acceleration equations to the free fall expressions.

Free Fall Equations

({{v=(Subscript[v, 0])-g t}, {Δ y=(1/2)((Subscript[v, 0])+v) t}, {Δ y=(Subscript[v, 0]) t-(1/2)g (t^2)}, {(v^2)=(Subscript[v, 0]^2)-2 g Δ y}})

Example . - A Dropped Ball

Solution

h=6.00(QuantityUnit)  and  g=9.80(QuantityUnit)/(((QuantityUnit))^2)

Δ y=-h , (Subscript[v, 0])=0  and  t=?

Δ y=(Subscript[v, 0]) t-(1/2)g (t^2)  ⟹   -h=0-(1/2)g (t^2)

h=(1/2)g (t^2).

t=(Sqrt[((2 h)/g)])==(Sqrt[((2×6.00(QuantityUnit))/(9.80(QuantityUnit)/(((QuantityUnit))^2)))])=1.11(QuantityUnit)

Solution

(v^2)=(Subscript[v, 0]^2)-2 g Δ y

(v^2)=2 g h .

v=-(Sqrt[2 g h])=-(Sqrt[2 ×9.80(QuantityUnit)/(((QuantityUnit))^2)×6.00(QuantityUnit)])=-10.8(QuantityUnit)/(QuantityUnit)

Example . - pward Initial Velocity

Solution

(Subscript[v, 0])=30.0(QuantityUnit)/(QuantityUnit) , v=0 , (Subscript[y, max])=Δ y=?

(v^2)=(Subscript[v, 0]^2)-2 g Δ y   ⟹   (Subscript[y, max])=Δ y=((Subscript[v, 0]^2)/(2 g))=45.9(QuantityUnit)

Solution

Δ y=(Subscript[v, 0]) t-(1/2)g (t^2)   ⟹   0=t ((Subscript[v, 0])-(1/2)g t)   ⟹    t=0   and   (Subscript[v, 0])-(1/2)g t=0

t=((2 (Subscript[v, 0]))/g)=6.12(QuantityUnit)

Solution

v=(Subscript[v, 0])-g t   ⟹   0=(Subscript[v, 0])-g (Subscript[t, up])  ⟹   (Subscript[t, up])=((Subscript[v, 0])/g)=3.06(QuantityUnit).

Solution

Δ y=(Subscript[v, 0]) t-(1/2)g (t^2)   ⟹   35.0(QuantityUnit)=(30.0(QuantityUnit)/(QuantityUnit)) t-(1/2)(9.80(QuantityUnit)/(((QuantityUnit))^2)) (t^2)

t=1.57(QuantityUnit)  and  t=4.55(QuantityUnit)