SME
Calculus I · Problem type
Displacement and distance from velocity
Use this for “A particle moves with v(t) = 3t − 6 m/s for 0 ≤ t ≤ 4 s. Find its displacement and the total distance it travels.”
Chart
v = v₀ + aT = (−4) + 2 × 5 = 6 m/s
Δx
=
v₀ + v
2
×
T
Δx
=
(−4) + 6
2
×
5
=
5
m
It turns round at 2 s, where the velocity is 0.
Area above the axis −4 m, below 9 m: displacement 5 m, distance travelled 13 m.
This is an example. Type your own values to start a new problem.
v(t)
=
v₀ (m/s)
+
a (m/s²)
×
t
Δx
=
∫
0
v(t)
dt
=
Δx (m)
D
=
∫
0
|v(t)|
dt
=
D (m)
v
Velocity at the end, t = T
worked out (not typed) · m/s
tₜᵤᵣₙ
Time when the velocity is 0
worked out (not typed) · s
Formulas
v
=
v₀
+
a
×
T
6 = −4 + 2 × 5
tₜᵤᵣₙ
=
−
v₀
a
2
=
−
−4
2
Δx
=
v₀
×
T
+
1
2
×
a
×
T
2
5
=
−4
×
5
+
1
2
×
2
×
5
2
D
=
|
v₀
×
tₜᵤᵣₙ
2
|
+
|
v
×
T
− tₜᵤᵣₙ
2
|
13
=
|−4
×
2
2
|
+
|6
×
5 − 2
2
|
Assumptions
•
The velocity is v(t) = v₀ + at from t = 0 to t = T. It is signed: + one way along the line, − the other.
•
Displacement is the integral of v from 0 to T. An antiderivative of v₀ + at is v₀t + ½at², so Δx = v₀T + ½aT².
•
Distance travelled is the integral of |v|, so every stretch counts as positive. Split the integral where v = 0, at tₜᵤᵣₙ = −v₀ ÷ a, and add the sizes of the pieces.
•
If tₜᵤᵣₙ is not between 0 and T, the object never turns round during the trip, and the distance is |Δx|.
Step-by-step
WE KNOW
v₀ = −4 m/s, a = 2 m/s², T = 5 s
FIND
velocity at the end, t = T (v), time when the velocity is 0 (tₜᵤᵣₙ), displacement, the integral of v (Δx), distance travelled, the integral of |v| (D)
Step 1 · Find velocity at the end, t = T (v)
Use v = v₀ + a × T
The
velocity
changes
by
a
each
second:
add
a
×
T
to
the
starting
velocity.
v
=
−4
+
2
×
5
v
=
−4
+
10
v
=
6
m/s
Step 2 · Find time when the velocity is 0 (tₜᵤᵣₙ)
Use tₜᵤᵣₙ = −v₀ ÷ a
Set
v₀
+
at
=
0
and
solve
for
t:
the
moment
the
velocity
passes
through
0.
tₜᵤᵣₙ
=
−
−4
2
tₜᵤᵣₙ = 2 s
Step 3 · Find displacement, the integral of v (Δx)
Use Δx = v₀ × T + ½ × a × T²
Integrate
v
(t)
=
2t
−
4
from
0
to
T
:
an
antiderivative
is
t
2
−
4t,
and
it
is
0
at
t
=
0.
Δx
=
−4
×
5
+
1
2
×
2
×
5
2
5
∫
0
(2
t
− 4) dt
=
[
t
2
−
4t]
from
0
to
5
=
5
Δx = 5 m
Step 4 · Find distance travelled, the integral of |v| (D)
Use D = |v₀ × tₜᵤᵣₙ ÷ 2| + |v × (T − tₜᵤᵣₙ) ÷ 2|
Integrate
v
from
0
to
tₜᵤᵣₙ
and
from
tₜᵤᵣₙ
to
T
.
One
piece
is
negative,
so
add
their
sizes:
distance
counts
both
ways
as
positive.
0 < 2 < 5: it turns round during the trip
D
=
|−4
×
2
2
|
+
|6
×
5 − 2
2
|
2
∫
0
(2
t
− 4) dt
=
[
t
2
−
4t]
from
0
to
2
=
−4
5
∫
2
(2
t
− 4) dt
=
[
t
2
−
4t]
from
2
to
5
=
9
D
=
|−4|
+
|9|
D
=
13
m
Check
Put the numbers back in. Both sides should match.
6 = −4 + 2 × 5 ✓
2
=
−
−4
2
✓
5
=
−4
×
5
+
1
2
×
2
×
5
2
✓
13
=
|−4
×
2
2
|
+
|6
×
5 − 2
2
|
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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