SME
University Physics I: Mechanics
Kinematics
Use this for “A car at 5 m/s speeds up at 2 m/s² for 4 s. How fast is it going, and how far has it gone?”
Chart
v = v₀ + at = 5 + 2 × 4 = 13 m/s
Δx
=
v₀ + v
2
×
t
=
5 + 13
2
×
4
Δx = 36 m
The shaded area above the axis is the displacement (+).
This is an example. Type your own values to start a new problem.
v (m/s)
=
v₀ (m/s)
+
a (m/s²)
t (s)
Δx
Displacement
calculated
Formulas
v
=
v₀
+
a
×
t
13 = 5 + 2 × 4
Δx
=
v₀
×
t
+
1
2
×
a
×
t
2
36
=
5
×
4
+
1
2
×
2
×
4
2
v
2
=
v₀
2
+
2
×
a
×
Δx
13
2
=
5
2
+
2
×
2
×
36
Δx
=
1
2
×
(
v₀
+
v
)
×
t
36
=
1
2
×
(5
+
13)
×
4
Assumptions
•
Acceleration is constant, and the clock starts (t = 0) when the velocity is v₀.
•
Motion is along a straight line. Pick a positive direction; signs follow it.
•
Free fall: a = −9.81 m/s² (−32.2 ft/s²) if up is positive.
•
Δx is displacement (change in position), the shaded area under the v–t line, not total distance traveled.
Step-by-step
WE KNOW
t = 4 s, v₀ = 5 m/s, a = 2 m/s²
FIND
final velocity (v), displacement (Δx)
Step 1 · Find final velocity (v)
Use v = v₀ + a × t
Velocity
changes
by
a
every
second,
so
add
a
·
t
to
the
starting
velocity.
v
=
5
+
2
×
4
v
=
5
+
8
v
=
13
m/s
Step 2 · Find displacement (Δx)
Use Δx = v₀ × t + ½ × a × t²
Add the distance covered at the starting velocity to the extra distance from accelerating.
Δx
=
5
×
4
+
1
2
×
2
×
4
2
Δx
=
5
×
4
+
1
2
×
2
×
16
Δx = 20 + 16
Δx = 36 m
Check
Put the numbers back in. Both sides should match.
13 = 5 + 2 × 4 ✓
36
=
5
×
4
+
1
2
×
2
×
4
2
✓
13
2
=
5
2
+
2
×
2
×
36
✓
36
=
1
2
×
(5
+
13)
×
4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Velocity and acceleration from x(t)
Use this for “x = 2t³ − 9t² + 12t + 4 (m, s). Find the velocity and acceleration at t = 3 s.”
A projectile’s position and velocity at a time
Use this for “A ball is thrown from 2 m up at 20 m/s, 60° above level. Where is it 2.5 s later, and how fast and which way is it moving?”