Physics
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Speed
…that’s how fast I am
going…
SCALAR
Velocity
…that’s my speed and
direction…
VECTOR
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
How can we illustrate motion graphically?
Position vs Time graphs
_Very useful for rectilinear motion for they show the way
in which motion took place (continuous or not, slow, fast,
constant or not, etc)
_Position (d) is the y-axis, whereas time (t) is the x-axis
_Slope gives you speed
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Description of motion
( Position vs time graph):
_60 m in 10 sec
_stops (0 m) for 5 sec
_100 m in 25 sec (opposite
direction = negative
Is it possible to have a
direction of axis)
vertical section in this
_40 m in 15 sec
graph? Why?
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Description of motion
( Position vs time graph):
_
12 m in 4 sec
_
0 m in 4 sec
_
12 m in 12 sec
(opposite direction)
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Practice Exercises:
Page 3.10 – 3.11, Ex 3.2 (b, d)
(use “a” as an example to draw the trajectory)
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
What is average speed?
Average speed =
total distance
time interval
Mathematically:
𝑣𝑎𝑣𝑒
∆𝑑
=
∆𝑡
∆𝑑 : distance between initial and final position
∆𝑡 : time required to travel said distance
𝑣𝑎𝑣𝑒 : average speed
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Speed and unit conversion
_Car travelling at 50 m/s, how much is the speed
in km/h?
𝟓𝟎
𝟏𝟎𝟎𝟎
𝟏
𝟑𝟔𝟎𝟎
=
𝟓𝟎 ∗𝟑𝟔𝟎𝟎
𝟏 ∗𝟏𝟎𝟎𝟎
= 180 km / h
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Examples of calculations (p. 3.19):
At an F1 race held on a 4.85 km circuit JV’s car completes 50th
lap in 1 h 12 min 15 s. Knowing that the previous one ended at
1h 10 min 45 s, calculate his average speed for the 50th lap.
𝑣𝑎𝑣𝑒 =
𝑣𝑎𝑣𝑒
∆𝑑
∆𝑡
4.85 𝑘𝑚
=
0.025ℎ
𝑣𝑎𝑣𝑒 = 194 𝑘𝑚/ℎ
∆𝑡 = t2 - t1 = (135s – 45s) = 90s
(90s / 3600s = 0.025h)
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Practice Exercises:
Page 3.19, Ex 3.9
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Speed (scalar)
Velocity (vector)
Ratio of total distance and
time interval
Ratio of total displacement
and time interval
Right or Wrong?
_v = 3km/h
_v = 7 m/s, 20°
Right or Wrong?
_ = 3km/h
_
= 7 m/s, 20°
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Parallel vectors
An airport moving sidewalk
moves with velocity of 4 km/h
with respect to the floor. A man
walks with velocity of 3 km/h
with respect to moving
sidewalk. What’s his velocity
with respect to the floor?
Vt = Vs + Vm (vector symbol)
Vt = 7 km/h
Opposite vectors
An airport moving sidewalk
moves with velocity of 4 km/h
with respect to the floor. A
child walks with velocity of 6
km/h with respect to moving
sidewalk, in opposite direction
to it. What’s his velocity with
respect to the floor?
Vt = Vc - Vs (vector symbol)
Vt = 2 km/h
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Practice Exercises:
Page 3.30, Ex 3.18 (info on 3.17)
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Uniform rectilinear motion: constant velocity
(constant speed and direction)
_Simplest motion
_Limited duration
_Position vs Time graphs will provide info on speed,
therefore velocity.
_Positive slopes, positive axis direction (vice verse)
_Slope = velocity
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
P
V
t
t
V
P
t
t
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Practice Exercises:
Page 3.45, Ex 3.29
Page 3.47, Ex 3.31
Page 3.49, Ex 3.32
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
We can find velocity using a P vs t graph for uniform rectilinear
motion. But, can we find position using a V vs t graph for uniform
rectilinear motion?
YES! Calculating the “area under the curve” for V vs t graphs
_See Fig. 3.14 a and 3.14b, p. 3.50
_3.14a (P vs t) train travelled 2m in 5 s
_3.14b (V vs t) at 5s, perpendicular line to cut graph.
_3.14b Rectangle (or square) formed
_Calculating area of ABCD implies V * t (m/s * s), so we obtained
position (in meters)
_For 5 s (V vs t), we get d = (5s * 0.4m/s) d = 2m (check 3.14a!)
_Try for 3s now!
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
In uniform rectilinear motion, the area under the curve
of a V vs t graph provides not only the value of distance
(at any given time), but also displacement since distance
and displacement are equal for this type of motion
*Displacement and distance travelled are both equal and
calculated as the area under the curve for any type of
rectilinear movement (no change in direction)
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
When direction changes:
_portion of area under the curve on positive side of
axis (positive value)
_portion of area under the curve on negative side of
axis (negative value)
_distance travelled is then the sum of the absolute
values of either area under the curve
_displacement is the algebraic sum of said areas
under the curve
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
_0 to 3s, vΔt = 1m/s * 3s = 3m
_3 to 6s, vΔt = 2m/s * 3s = 6m
_6 to 8s, vΔt = 1m/s * 2s = 2m
_8 to 9s, vΔt = 0m/s * 1s = 0m
_9 to 10s, vΔt = -2m/s * 1s = -2m
_10 to 11s, vΔt = -3m/s * 1s = -3m
_11 to 15s, vΔt = -4m/s * 4s = -16m
Distance travelled: 3m+6m+2m+(2m)+(3m)+(16m) = 32m
Displacement: (3+6+2)m – (2+3+16)m = -10m
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Practice Exercises:
Page 3.54, Ex 3.34
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Motion Equation:
v = Δd / Δt
In many cases t1 = 0, so
d2 = v t2 + d1
Δd = v * Δt
d2 – d1 = v * Δt
d2 = vΔt + d1
or just:
d = v t + d1
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
A cyclist travels at 12 m/s in uniform rectilinear
motion. We begin observing his motion when he
is 250m from start point.
a) Motion equation?
d = 12t + 250
b) Position of cyclist at 10s of observation
d = 12(10) + 250
d = 370m
c) When will he be 1000m from start line?
d = 12t + 250
t = (d – 250) / 12
t = 62.5s
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Practice Exercises:
Page 3.57, Ex 3.35
PHS 5042-2 Kinematics & Momentum
Speed & Velocity
Practice exercises:
Page 3.100 – 3.104,
Ex 3.52 – 3.56 and 3.58