Motion equations will be developed in class.

1. Lesson 4Lesson 4
Equations of MotionEquations of Motion
Nelson Reference Pages 36-39Nelson Reference Pages 36-39
Note: Vector arrows have beenNote: Vector arrows have been
dropped on vector quantities anddropped on vector quantities and
the vectors have been boldedthe vectors have been bolded
instead.instead.

2.  For the upper graph, we
want to find ΔdT , { the
total displacement}
 We know this is equal to
the ____________ .
 One way to calculate
this area is to use the
average velocity (vAVG).
The equation is as
shown and the shaded
area represents the total
displacement. We could
also write the equation
as:
ΔdT = ½(v1+v2)Δt
t i m e
v 2
v v s t
v 1
t 1
v A V G
v A V G
= ( v 1
+ v 2
) / 2
t = t 2
- t 1
t 2t
t i m e
v 2
v v s t
v 1
t 1
v A V G
t 2t
d T
= v A V G
t

3.  From this graph, we canFrom this graph, we can
see that:see that: vv22 == vv11 ++ ΔΔvv
 But we know that:But we know that:
aa = Δ= Δv /v / Δt,Δt, so rearrangingso rearranging
ΔΔvv == aa ΔtΔt and we canand we can
rewrite the equation as:rewrite the equation as:
vv22 == vv11 ++ aaΔtΔt or we canor we can
algebraically rearrangealgebraically rearrange
this equation to get:this equation to get:
vv11 == vv22 -- aaΔtΔt
tim e
v 2
v v s t
v 1
t2
t1
v = v 2
- v 1
v
}
t

4. Last Three EquationsLast Three Equations
 We will use theWe will use the vv-t-t
graph to derive thegraph to derive the
remaining tworemaining two
displacementdisplacement
equations on theequations on the
chalk board.chalk board.
 The last equation willThe last equation will
be derivedbe derived
algebraically.algebraically.
t im e
v 2
v v s t
v 1
t 1 t 2t
A 2
A 1
A 3
A 4
A 4 = A 1 + A 2 + A 3
( A 4 IS T H E B IG R E C T A N G L E )

5. Chalkboard ExampleChalkboard Example
 Jason is jogging east with a constantJason is jogging east with a constant
speed ofspeed of 3.0 m/s3.0 m/s. Trevor is. Trevor is 40 m40 m behindbehind
Jason and is cycling atJason and is cycling at 2.0 m/s [E].2.0 m/s [E]. IfIf
Trevor starts to accelerate atTrevor starts to accelerate at 0.25 m/s0.25 m/s22
[E][E]
when he iswhen he is 40 m40 m behind, how long (inbehind, how long (in
seconds) will it take Trevor to catch up?seconds) will it take Trevor to catch up?
 Ans. 22 sAns. 22 s

6. Chalkboard ExampleChalkboard Example
 Tony is driving atTony is driving at 130 km/h [W]130 km/h [W] when hewhen he
passes a non-moving police car. It takespasses a non-moving police car. It takes
the officerthe officer 4.0 s4.0 s to react, he then starts toto react, he then starts to
accelerate ataccelerate at 4.0 m/s4.0 m/s22
[W][W] forfor 10 s10 s andand
then he moves with uniform motion untilthen he moves with uniform motion until
he catches up to Tony.he catches up to Tony.
 Do the following:Do the following:
 Draw aDraw a vv-t graph. Let t = 0.0 s when Tony-t graph. Let t = 0.0 s when Tony
just starts to pass the officer.just starts to pass the officer.
 Determine the amount of time it takes theDetermine the amount of time it takes the
officer to just catch up to Tony.officer to just catch up to Tony. Ans. 92 sAns. 92 s

7. Practice QuestionsPractice Questions
 Nelson TB:
 Page 39 #1 - 4
 Page 54 #39, 40, 55, 57, 59
From McGraw-Hill TB
A car starts from rest and travels a linear distance
of 5.0 x101
m in a time of 6.0 s
a.) Determine v2 at 6.0s (Ans +16.7 m/s)
b.) Determine acceleration. (Ans +2.8 m/s2
)