UNIT 1: 1-D KINEMATICS

Lesson 5:

Kinematic Equations

CENTRE HIGH: PHYSICS 20

Reading Segment:

Kinematic Equations

To prepare for this section, please read:

Unit 1: p.19

Kinematic Equations

When an object has a constant acceleration,

there are many equations which describe the motion.

A complete list of equations is on the next page

(and on your formula sheet)

Here is a list of all the equations:

a = vf - vi vf2 = vi

2 + 2 a d

t

d = vi t + 0.5 a t2 d = vf t - 0.5 a t2

d = vi + vf t

2

where vavg = vi + vf

2

To figure out which equation to use, you make a list of

the 5 kinematic variables:

vi, vf, a, d, t

If you can identify the variable that is NOT involved

in the question, you can choose the proper equation.

This will be illustrated in the next pages.

Variables involved in question:

vi, vf, a, t

Variable NOT involved in the question:

d

Equation to use: a = vf - vi

t

This is the only kinematic equation that

does not have displacement (d) in it.

Variables involved in question:

vi, vf, a, d

Variable NOT involved in the question:

t

Equation to use: vf2 = vi

2 + 2 a d

This is the only kinematic equation that

does not have time (t) in it.

The process is similar for the other equations:

Equations Variables

vi vf a d t

d = vi t + 0.5 a t2

d = vf t - 0.5 a t2

d = vi + vf t

2

Note:

If the object has a constant velocity:

- the acceleration is zero

- the final velocity is equal to the initial velocity (vi = vf)

Use the equation: v = d

t

Ex. 1 A ball is rolled up a constant incline at a speed of 9.0 m/s.

How much time is required to reach a maximum

displacement of 6.45 m ?

Diagram: Rest

9.0 m/s 6.45 m

If it has reached its maximum displacement,

then it cannot go any higher.

Thus, it has come to rest (vf = 0)

List: Ref: Up slope is positive

vi = +9.0 m/s Down is negative

vf = 0

a d = +6.45 m

t ?

Make a list of the kinematic variables.

Be certain you get the right sign for the vector quantities.

List: Ref: Up slope is positive

vi = +9.0 m/s Down is negative

vf = 0

a d = +6.45 m

t ?

d = vi + vf t

2

This is the only equation that does not

have acceleration (a) in it.

d = vi + vf t

2

d = vi + vf t

2

d = vi t

2

Since vf = 0, then remove this variable from the equation.

d = vi t

2

2 d = vi t

t = 2 d = 2 (6.45 m)

vi 9.0 m/s

= 1.4 s

Ex. 2 A particle experiences an acceleration of 470 m/s2 North

for a total time of 960 milliseconds. If its displacement

is 0.123 km South, then find the initial velocity.

List: Ref: North is positive

vi = ? South is negative

vf

a = +470 m/s2

d = -0.123 103 m = -123 m

t = 960 10-3 s = 0.960 s

Make a list of the kinematic variables.

Be certain you get the right sign for the vector quantities

and all quantities are in standard units.

List: Ref: North is positive

vi = ? South is negative

vf

a = +470 m/s2

d = -0.123 103 m = -123 m

t = 960 10-3 s = 0.960 s

Equation: d = vi t + 0.5 a t2

This is the only equation that does not have

the final velocity (vf) in it.

d = vi t + 0.5 a t2

d - 0.5 a t2 = vi t

vi = d - 0.5 a t2

t

= (-123 m) - 0.5 (470 m/s2) (0.960 s)2

0.960 s

= -354 m/s

vi = 354 m/s South

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 1 p. 10 #1 - 10