SME
Grade 11 Science · Physics
Motion in one dimension: displacement, velocity and acceleration
Use this for “A cart starts at 4 m/s and speeds up at 2.5 m/s² for 6 s. How fast is it going, and how far did it travel?”
Chart
v = v₀ + at = 4 + 2.5 × 6 = 19 m/s
Δx
=
v₀ + v
2
×
t
=
4 + 19
2
×
6
Δx = 69 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 (m)
=
v₀ (m/s)
×
t (s)
+
½
×
a (m/s²)
×
(
t (s)
)
2
Formulas
v
=
v₀
+
a
×
t
19 = 4 + 2.5 × 6
Δx
=
v₀
×
t
+
1
2
×
a
×
t
2
69
=
4
×
6
+
1
2
×
2.5
×
6
2
v
2
=
v₀
2
+
2
×
a
×
Δx
19
2
=
4
2
+
2
×
2.5
×
69
Assumptions
•
The acceleration is constant the whole time.
•
Choose which way is +; an object slows down when a and v point opposite ways.
•
Displacement Δx is where it ends up from the start, not how far it went.
•
The shaded area under the v–t line is the displacement.
•
v² = v₀² + 2aΔx needs no time: use it when the time is not given.
Step-by-step
WE KNOW
v₀ = 4 m/s, a = 2.5 m/s², t = 6 s
FIND
final velocity (v), displacement (Δx)
Step 1 · Find final velocity (v)
Use v = v₀ + a × t
Start
at
v₀
and
add
a
for
every
second.
v
=
4
+
2.5
×
6
v
=
4
+
15
v
=
19
m/s
Step 2 · Find displacement (Δx)
Use Δx = v₀ × t + ½ × a × t²
The distance at the starting velocity, plus the extra that the steady acceleration adds.
Δx
=
4
×
6
+
1
2
×
2.5
×
6
2
Δx
=
4
×
6
+
1
2
×
2.5
×
36
Δx = 24 + 1.25 × 36
Δx = 24 + 45
Δx = 69 m
Check
Put the numbers back in. Both sides should match.
19 = 4 + 2.5 × 6 ✓
69
=
4
×
6
+
1
2
×
2.5
×
6
2
✓
19
2
=
4
2
+
2
×
2.5
×
69
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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