PHYSICS 1: 1D Kinematics
Kalamazoo Valley Community College
Phy 112
• Kinematics is a way of
tracking the motion
of an object.
• There are five
variables for a
direction used for
most kinematic
equations.
• Three variables are
required to solve any
problem.
• Use a table to
organize your
information.
X-direction
Δ𝑥
ax
V0x
Vfx
time
y-direction
Δ𝑦
ay
V0y
vfy
time
1D Example
A car starts from rest and accelerates to a speed
of 18.0m/s in 50.0m. Solve for the time taken
and the acceleration of the car.
• The solution starts by finding key
information in the problem statement.
• Look for keywords containing important
information.
• Make a table to fill in known variables.
1D Example Continued
A car starts from rest and accelerates to a speed
of 18.0m/s in 50.0m. Solve for the time taken
and the acceleration of the car.
X-direction
Δ𝑥
50.0m
ax
???
V0x
0
Vfy
18.0m/s
time
???
1D Example
Three variables are
required to solve for
the others. There is
an exception to this
that will be covered
later.
We need to pick an
equation that can
solve for one of the
missing varibles.
X-direction
Δ𝑥
50.0m
ax
???
V0x
0
Vfx
18.0m/s
time
???
Δ𝑑 = 𝑣0 ∙ 𝑡 + 12∙a ∙ 𝑡 2
𝑣𝑓 = 𝑣0 + a ∙ 𝑡
𝑣𝑓2 = 𝑣02 + 2 ∙ a ∙ Δ𝑑
𝑣𝑓2 = 𝑣02 + 2 ∙ a ∙ Δ𝑥
1D Example
2
2 + 2 ∙ a ∙ 50𝑚
18𝑚
=
0
𝑠
𝑚2
324 𝑠2
X-direction
Δ𝑥
50.0m
ax
???
V0x
0
Vfx
18.0m/s
time
???
Δ𝑥 = 𝑣0 ∙ 𝑡 + 12∙a ∙ 𝑡 2
𝑣𝑓 = 𝑣0 + a ∙ 𝑡
𝑣𝑓2 = 𝑣02 + 2 ∙ a ∙ Δ𝑥
100𝑚
=
02
a ∙ 100𝑚
+
100𝑚
𝑎 = 3.24𝑠𝑚2
Δ𝑥 = 𝑣0 ∙ 𝑡 + 12∙a ∙ 𝑡 2
50𝑚 = 0 ∙ 𝑡 + 12∙3.24𝑠𝑚2 ∙ 𝑡 2
50𝑚 = 0 ∙ 𝑡 + 1.62𝑠𝑚2 ∙ 𝑡 2
𝑡=±
50𝑚
𝑚
1.62 2
𝑠
𝑡 = +5.56s