Definitions of the kinematic variables

Derivation of the Kinematic Equations - 1

Derivation of the Kinematic Equations - 2

Derivation of the Kinematic Equations - 3

Derivation of the Kinematic Equations - 4

Graphs of the Kinematic Equations

PROBLEM-SOLVING STRATEGY 2.1 Motion with constant acceleration

S ET U P

1. You must decide at the beginning of a problem where the origin of coordi-nates is and which axis direction is positive. The choices are usually a mat-ter of convenience; it is often easiest to place the origin at the object’slocation at time then Draw a diagram showing your choices.Then sketch any relevant later positions of the object in the same diagram.

2. Once you have chosen the positive axis direction, the positive directions forvelocity and acceleration are also determined. It would be wrong to define xas positive to the right of the origin and velocities as positive toward the left.

3. It often helps to restate the problem in prose first and then translate theprose description into symbols and equations. When (i.e, at what value of t)does the particle arrive at a certain point? Where is the particle when itsvelocity has a certain value? (That is, what is the value of x when has thespecified value?) The next example asks, “Where is the motorcyclist whenhis velocity is Translated into symbols, this becomes “What is thevalue of x when

4. Either on your diagram or in a list, write known and unknown quantitiessuch as x, and Write in the values for those that are known.Be on the lookout for implicit information; for example, “A car sits at astoplight” usually means that and so on.

SO LV E

5. Once you’ve identified the unknowns, you may be able to choose an equa-tion from among Equations 2.8, 2.12, 2.13, and 2.14 that contains only oneof the unknowns. Solve for that unknown; then substitute the known valuesand compute the value of the unknown. Carry the units of the quantitiesalong with your calculations as an added consistency check.

R E F L EC T

6. Take a hard look at your results to see whether they make sense. Are theywithin the general range of magnitudes you expected? If you change one ofthe given quantities, do the results change in a way you can predict?

v0x 5 0,

ax .v0x ,vx ,x0 ,

vx 5 25 m/s?”25 m/s?”

vx

x0 5 0.t 5 0;

There are 12 worked out example problems in chapter 2 of the textbook (in blue boxes). You must work your way through each of them to make sure you understand the rationale physicists employ in order to solve problems. In addition to getting an introduction to the laws that govern the physical universe, the other goal of this course is for the each student to develop reasoning skills in order to become a proficient problem solver.

Example 1: A go-cart is travelling around a 100 meter track. As it reaches the final lap, it is travelling at 5 m/s. Midway through that final lap it experiences a mechanical failure and is subjected to a constant acceleration of -0.2 m/s2. How long does it take to complete the final lap?

Solution: Recall that the kinematic equations are only valid for constant acceleration, so the trick here is to break the problem into two separate parts, each of which has a constant acceleration.

For first half of the final lap: xf – xo = 50 meters vo = 5 m/s a = 0 m/s2

let t1 be the time to complete the first half of the final lap

xf – xo = ½ a t12 + vo t1 so t1 = (xf – xo ) / vo

or t1 = (50 meters) / (5 m/s) = 10 seconds

For the second half we have: xf – xo = 50 meters vo = 5 m/s a = -0.2 m/s2

let t2 be the time to complete the second half of the final lap

We’ll solve this problem in two distinct ways:

A) Using vf2 – vo

2 = 2 a (xf – xo ) lets solve for vf , vf = vo2 + 2 a (xf – xo )

so vf = (5 m/s )2 + 2 (-0.2 m/s2 ) *(50 m) = 2.23 m/s (check units here), this is the speed it has when it finishes the second 50 meters of the final lap. With vf now known, we can determine the amount of time (t2) it took to slow to this speed. Using

vf = a * t2 + vo , we’ll solve for t2, t2 = (vf – vo) / a = (2.23 m/s – 5 m/s)/(-0.2 m/s2) so t2 = 13.8 seconds, yielding a total time of t1 + t2 = 10 + 13.8 = 23.8 seconds

B) Another solution to this problem would be to solve it in one step by using the quadratic equation. We have xf – xo = 50 meters vo = 5 m/s a = -0.2 m/s2 so in numbers we can write xf – xo = 0.5 a t22 + vo t2 as

-0.1 t22 + 5 t2 - 50 =0 , where for simplicity I‘ve dropped the units here. (You’ll need to be extra careful whenever you use the quadratic equation).

Solutions are t2 = 13.8 seconds, and 36.2 seconds. These are both mathematical solutions to this equation, but only one can be the actual solution to the real world problem. Having determined previously that the correct answer is 13.8 seconds, it is instructive to see how the other solution comes about.

We know that given the mechanical failure of the go-cart, it will fairly quickly come to rest. To determine how long that takes we’ll use the following equation:

vf = a * t2 + vo , and solve for the time t2 when vf = 0 m/s, then t2 = (vf – vo) / a = (0 m/s – 5 m/s)/(-0.2 m/s2) = 25 seconds

Clearly the go-cart would come to rest after 25 seconds, however our mathematical equation doesn’t impose a time when it is no longer valid. We assumed a constant acceleration without time limit. In the real world once the go-cart came to rest, it will remain at rest. However our equation has a constant acceleration of -0.2 m/s2. So apparently at 13.8 seconds it first crosses the finish line. At 25 seconds it comes to rest momentarily, before it reverses its direction, and at 36.2 seconds it crosses the finish line going in the opposite direction. Notice that it takes (25 -13.8) = 11.2 seconds from the time it first crosses the finish line to come to rest. It also takes 11.2 seconds (36.2 -25) to travel from its resting spot back to the finish line. Moral of the story: make sure you understand the conditions in which your equations you are using are valid!