The Kinematic Equations

Kinematics - (describing how things move)

Scalar (no direction) Vector (w/ direction)

Distance (d) Displacement (d)

Speed (s) Velocity (v)

Acceleration (a)

How far you travel Change in position

(How far you travel in a

given direction)

How fast you travel How fast you travel (in

a given direction)

Rate of change of velocity

Describing Motion

There are lots of different ways to describe

motion….

1. Words

2. Sketches

3. Time elapsed photographs

4. Physical Expressions (Equations)

5. Graphical Representation

KinematicsDescribes motion without regard to what causes it.

Uses equations to represent the motion of an object in terms of acceleration (a), initial velocity (vi), final velocity (vf), displacement (Δx) and time (t).

These five quantities are related by a group of equations that we call the BIG FOUR.

KINEMATIC EQUATIONS

Whenever possible, it will be convenient to place the frame of reference at the origin xi = 0 m when t0 = 0 s.

With this assumption, the displacement

Δx = xf – xi becomes

Δx = x.

Big 4 Equations

vf = vi + at v = ΔxΔt

Δx = vit + ½at2 vf2 = vi

2 + 2aΔx

KINEMATICS BIG FOUR

In BIG FOUR, the average velocity is simply the average of the initial velocity and the final velocity:

v = ½(vi + vf).

(This is a consequence of the fact that the acceleration is constant.)

KINEMATICS BIG FOUREach of the BIG FOUR equations is missing one of the five fundamental quantities.

The way you decide which of equation to use when solving a problem is to determine which of the fundamental quantities is missing from the problem –that is, which quantity is neither given nor asked for – and then use the equation that doesn’t have that variable.

KINEMATICS BIG FOUR

For example, if the problem never mentions the final velocity …

… vf is neither given nor asked for …

… the equation to use is the one that’s missing vf …

Δx = vit + ½at2

Example The Displacement of a Speedboat

The speedboat in the figure has a constant acceleration of +2.0 m/s2. If

the initial velocity of the boat is +6.0 m/s, find its displacement after

8.0 seconds.

Example The Displacement of a Speedboat

Reasoning Numerical values for the three unknown variables are listed in the data table. We’re asked to determine the displacement x of the speedboat, so it gets the question mark.

We choose Δx = vit + ½at2

Δx = (6.0 m/s)(8.0 s) + ½(+2.0 m/s2)(8.0 s)2

Δx = 48 m + 64 m = 112 m

x a vf vi t

? +2.0 m/s2 + 6.0 m/s 8.0 s

Example Catapulting a Jet

Reasoning The data are as follows:

The initial velocity vi = 0, since the jet starts from rest. The displacement x of the aircraft can be obtained from Δx = (vav)t, if we can determine the time t during which the plane is being accelerated.

But t is controlled by the value of the acceleration. With larger accelerations, jet reaches its final velocity in shorter times, as can be

seen by solving vf = vi + at for t.

x a vf vi t

? +31 m/s2 +62 m/s 0 m/s

Example Catapulting a Jet

Solving for t,

vf = vi + at

62 m/s = 0 m/s + 31 m/s2 (t)t = 2.0 s

Since the time is now known, the displacement can be found by using Δx = vi t + 1/2a(t)2 :

= (0) (2) + ½ (31 m/s2) (2.0 s)2 = +62 m

If a car’s initial velocity is +25 m/s, and it accelerated at a rate of +7.5 m/s2 over a period of 8.0 seconds what is the car’s final velocity?

vf = vi + at

vf = 25 m/s + (7.5 m/s2) (8.0 s)

vf = 85 m/s

x a vf vi t

+7.5 m/s2 ? + 25 m/s 8.0 s

If a kicker boots the ball at a velocity of +45 m/s and it is aided by a gust of wind to accelerate at +2.5 m/s2 and it stays in the air for 3.5 seconds. How far will the ball travel?

Δx = (45 m/s) (3.5 s) + ½ (2.5 m/s2) (3.5 s)2

Δx = +173 m

x a vf vi t

? 2.5 m/s2 + 45 m/s 3.5 s

A boy sledding down a hill has an initial speed of +12m/s. He continues to speed up and reaches a final velocity of +18m/s after traveling for 12 seconds. What distance does the boy travel.

v = Δx/Δt, so Δx = vav (t)

Δx = (vf + vi) (t) = (18 + 12) (12 s)

2 2

Δx = 180 m

x a vf vi t

? +18 m/s + 12 m/s 12 s

Mr. Billante is notorious for his road rage. After being cut off he accelerates at a rate of +12.3 m/s2

for +455 m. As he approaches the man who cut him off his final velocity is a ludicrous +125 m/s. What was his initial velocity?

vf2 = vi

2 + 2aΔx

(125)2 = vi2 + 2 (12.3 m/s2) (455 m)

vi = +66.6 m/s

x a vf vi t

455 m +12.3 m/s2 +125 m/s ? 12 s