xx v tΔ = Δ

2120

0

2 20

2y

2

In the case of free fall a becomes g = -9.81m/s

y y

y y y

y y y

y v t a t

v v a t

v v a y

Δ = Δ + Δ

= + Δ

= + Δ

2120

02 2

0 2

x x

x x x

x x x

x v t a tv v a t

v v a x

Δ = Δ + Δ= + Δ

= + Δ

How to Solve Kinematics ProblemsThere are several steps, an algorithm or recipe, needed to solve any kinematics problem.

1. READ THE PROBLEM CAREFULLY!!!! This means that you must ask yourself a lotof questions as you read, so you can interpret the situation described.

a. The first question you must ask, “What type of motion is this situation about;constant velocity or constant acceleration?”

Determining the answer is pretty easy. In the textbook the word acceleration ordeceleration often shows up which is a pretty good give away. However, I prefernot to do this with my problems as in the real world you may not be told soclearly. Instead I will often describe a situation where the velocity is one valueand later on some new value. If this “change in velocity” happens over a knowndistance OR known time then you have CONSTANT ACCELERATION.

Another important clue is if the motion appears to happen over a great distance ortime. If that is the case you have CONSTANT VELOCITY. In other wordsaccelerations in real life happen over short distances (maximum of a couple ofhundred metres) or short intervals (0 - 90 seconds).

Why is this question of the main importance? If your answer is constant velocitythen the only equation you may use to solve the problem is:

If your answer is constant acceleration then would not use the above equation butrather one or more of the following equations:

Of course the preceding comment is ignoring the mean speed theorem, but it isnot always advisable to do use this starting out.

Read the following practice exercise and determine what type of motion isdescribed, constant velocity or constant acceleration.

i. Practice 1: A jumbo jet flies from St. John’s to London Heathrow adistance of 4500 km at a steady 530 km/h. How long does this flight last?

(ANSWER at the end of the document)

ii. Practice 2: A jumbo jet takes off from St. John’s International Airport. Starting from rest it reaches its take off speed of 36.0 m/s over 1200metres of runway. How long does this take?

(ANSWER at the end of the document)

2. Next you need to draw a diagram listing the physical quantities you know and do notknow. The physical quantities are displacement, some times expressed as positions,interval times, initial velocity, final velocity and acceleration.

The diagrams are really just number lines with critical points noted. Such points includethe Start, End and any point in between where the motion changes or you knowadditional data.

Why a diagram and not just a “list of knowns and unknowns” as “we did in high school?” There are two main reasons. First the diagram can act as tool to help you reallyunderstand what is described in the problem. Secondly the list technique lends itself to“text book problems” not real world situations.

I will illustrate the previous examples.

i. Practice 1: A jumbo jet flies from St. John’s to London Heathrow adistance of 4500 km at a steady 530 km/h. How long does this flight last?

ii. Practice 2: A jumbo jet takes off from St. John’s International Airport. Starting from rest it reaches its take off speed of 36.0 m/s over 1200metres of runway. How long does this take?

3. Time to read again, but this time is not so bad. What is the first thing you are asked tofind, determine or solve for? We see that for both examples the interval time ()t) is thespecific quantity you are asked to solve for.

4. Knowing what you are looking for consult with your list of equations and find equationsthat contain this unknown. Then ask yourself a. if you know the rest of the values of the terms in that equation, then your set solve

the problem. b. if you do not know two or more terms see if there is another equation that you

know everything but the unknownc. if you cannot find such an equation this means you must do an intermediate

calculation to find another unknown, then try to determine the unknown youinitially set out to find.

Remember point (1a)? Your answer there will automatically restrict the equations youare allowed to use. Bear this in mind when selecting equations.

xx v tΔ = Δ

4500 8.49530

x

kmhx

x v tx kmt h

v

Δ = ΔΔ

Δ = = =

i. Practice 1: A jumbo jet flies from St. John’s to London Heathrow adistance of 4500 km at a steady 530 km/h. How long does this flight last?

Looking for )t and this is constant velocity. This means that I can only use the:

equation to solve this problem.

2120

0

x x

x x x

x v t a tv v a tΔ = Δ + Δ= + Δ

22

2

2 20

2 22 20

2

(35.0 0 )0.51

2 2(1200)

x x x

mx x s m

x s

v v a x

v vax m

= + Δ

−−= = =

Δ

2

0

0 (35.0 0) 68.570.51

x x x

msx x

mx s

v v a tv vt s

a

= + Δ− −

Δ = = =

ii. Practice 2: A jumbo jet takes off from St. John’s International Airport. Starting from rest it reaches its take off speed of 36.0 m/s over 1200metres of runway. How long does this take?

We are trying to find interval ()t). Consulting with the appropriate acceleration equations thereare only two that can solve for interval time.

While we know displacement ()x), initial velocity (v0x) and final velocity (vx), we don’t knowacceleration (ax). Therefore, we must first solve for acceleration using the other accelerationequation.

Now we can pick the easiest remaining equation to solve for interval.

Practice 1: A jumbo jet flies from St. John’s to London Heathrow a distance of 4500 km at asteady 530 km/h. How long does this flight last?

Constant velocity, because this trip is over a very large distance and should have a verylong duration. Real objects only accelerate for very short distances and intervals.

Practice 2: A jumbo jet takes off from St. John’s International Airport. Starting from rest itreaches its take off speed of 36.0 m/s over 1200 metres of runway. How long does this take?

Constant acceleration, the jet should start with no velocity but after travelling a shortdistance of 1200 metres has a positive velocity. This means velocity has changed whichis the event that defines acceleration!