2.5. The Kinematic Equations http://www.ck12.org
- Do the units of the area in the velocity-time plane provide any information concerning the rocket?
Answer:Yes. Using the same method to find the units of the area, multiplying the vertical units by the horizontal
units, we have
(m
s)(s
1)
=m. Thus, the area in the velocity-time plane represents the displacement of the rocket at a
specific time. Let’s use the area in the velocity-time plane in order to generate a position-time graph for the rocket.
Let’s say the initial position of the rocket is +10 m. Consider the trapezoidal area under the graph for the time interval
[0, 1] (seeFigure2.17). It is helpful to break the trapezoid into a rectangle and triangle in order to compute the area.
The rectangle can be formed with a “horizontal side” (length) of “1.0 second long” and a “vertical side” (height) of
“+20 m/s high”. The triangular area can be formed with a base of 1.0 second and height of +10 m/s. The area of the
rectangular is then( 1 .0 s)(20 m/s) = +20 m and the area of the triangle is^12 ( 1 .0 s)(+10 m/s) = +5 m. The total
area gives a displacement of +25 m. Recall however, that the rocket started at +10 m. Therefore, att= 1. 0 s, the
rocket’s position is(+ 10 )+(+ 25 ) = +35 m.
Next, look at the time interval [1, 2]. A rectangle with the length 1.0 s and the height +10 m/s, and a triangle with
a base of 1.0 s and a height of 10 m/s can be constructed. The area of the rectangle is +10 m and the area of the
triangle is again, +5 m. The position of the rocket att=2 s is therefore(+ 35 ) + (+ 10 ) + (+ 5. 0 ) = +50 m. The
interval [2, 3] is just composed of a triangle with the same position +5.0 m. Thus the rocket’s position att= 3 .0 s
is,(+ 50 )+(+ 5. 0 ) = +55 m. Notice that the velocity of the rocket is zero att= 3 .0 s. So, att= 3 .0 s, the rocket
has position +55 m (55 m above ground level), zero velocity, and an acceleration of−10 m/s^2. Remember, the
acceleration is the slope of the line in the velocity-time plane, and the slope is constant. Since the velocity of the
rocket is zero att= 3 .0 s, it suggests the rocket has reached its highest position and will begin its descent earthward.
If this is indeed the case then the position of the rocket att= 4 .0 s must be lower than 55 m. For the time interval [3,
4], we see that the triangle is the same as for the time interval [2, 3]. However, the velocity is a negative quantity for
the [3, 4] time interval, and therefore the position will be negative. Hence, instead of the +5 m we determined for the
interval [2, 3] we have -5.0 m. Since the rocket is at +55 m att= 3 .0 s, att=4 s, it is at(+ 55 )+(− 5 ) = +50 m.
See if you can determine that att= 5 .0 s, the rocket is at + 35 m and att= 6 .0 s, it is at +10 m. In other words, it is
in the same position att= 6 .0 s as when it was launched, att=0.
(Why do you think that is the case?)
Deriving the Kinematic Equations
We begin with a familiar problem.
Parachutists can land safely because they are not in free fall (more than the force of gravity acts upon them). Air
resistance early on in the jump is responsible for quickly bringing a parachutist to a comparatively small, constant
rate of fall, commonly called the terminal velocity. If the terminal velocity is 7 m/s how far will a parachutist fall in
3 seconds?
http://demonstrations.wolfram.com/MotionOfAParachuter/
We have:x=vt= 7 × 3 =21 m. Problems such as this are easy because the velocity is constant.
Consider a question with varying velocity.
A toy rocket is launched with an initial velocity of +30 m/s. How far has the rocket traveled after three seconds if it
started from an elevation of +10 m?
We can still solve this problem usingx=vtbut we need to determine what the appropriatevis. We make use of the
area in the velocity-time plane do to this.
Rather than add up the areas in the velocity-time plane, one second at a time, let’s find the area under the graph for
the interval of time [0, 3]. This area is a triangle, with a base of 3 s and a height of 30 m/s. The area of the triangle
is^12 ( 3. 0 )( 30 ) =45 m (remember we had an initial position of +10 m in the problem so we end up with +55 m). The
equation used to find 45 m can be reinterpreted as( 3. 0 )
[
( 30 )
2]
= 3. 0 ×15. Geometrically, this represents a rectangleof “length” 3 s and “height” 15 m/s. The average velocity for the time interval [0, 3.0] is(^30 + 20 )=15 m/s. This result