SME
Dynamics
Particle kinematics
Use this for “A particle moves along a line with s = 2t³ − 9t² + 12t + 5 (m, s). Find s, v and a at t = 3 s.”
Chart
s(3) = 14
Tangent at t = 3: slope s′(3) = 12, y = 12t − 22
This is an example. Type your own values to start a new problem.
s(t)
=
c₃
t³
+
(
c₂
)
t²
+
c₁
t
+
c₀
s(
t (s)
)
=
s (m)
v
Velocity
worked out (not typed) · m/s
a
Acceleration
worked out (not typed) · m/s²
Formulas
s
=
c₃
×
t
3
+
c₂
×
t
2
+
c₁
×
t
+
c₀
14
=
2
×
3
3
+
(−9)
×
3
2
+
12
×
3
+
5
v
=
3
×
c₃
×
t
2
+
2
×
c₂
×
t
+
c₁
12
=
3
×
2
×
3
2
+
2
×
(−9)
×
3
+
12
a
=
6
×
c₃
×
t
+
2
×
c₂
18 = 6 × 2 × 3 + 2 × (−9)
Assumptions
•
Motion along one straight line: s is the position from a fixed origin, + one way and − the other.
•
s(t) = c₃t³ + c₂t² + c₁t + c₀, with s in meters and t in seconds.
•
v = ds/dt and a = dv/dt, by the power rule term by term: tⁿ gives ntⁿ⁻¹.
•
Where v = 0 and changes sign, the particle turns round.
Step-by-step
WE KNOW
c₃ = 2 m/s³, c₂ = −9 m/s², c₁ = 12 m/s, c₀ = 5 m, t = 3 s
FIND
position (s), velocity (v), acceleration (a)
Step 1 · Find position (s)
Use s = c₃ × t³ + c₂ × t² + c₁ × t + c₀
Put
the
time
into
s
(
t
)
=
2
t
³
−
9
t
²
+
12
t
+
5,
one
term
at
a
time,
and
add.
s
=
2
×
3
3
+
(−9)
×
3
2
+
12
×
3
+
5
s
=
2
×
27
+
(−9)
×
9
+
12
×
3
+
5
s
=
54
+
(−81)
+
36
+
5
s
=
14
m
Step 2 · Find velocity (v)
Use v = 3 × c₃ × t² + 2 × c₂ × t + c₁
v
=
ds/dt:
each
cₙtⁿ
gives
ncₙtⁿ⁻¹
and
c₀
drops
out.
Put
t
in.
v
=
3
×
2
×
3
2
+
2
×
(−9)
×
3
+
12
d/dt
(
2
t
³
−
9
t
²
+
12
t
+
5)
=
6
t
²
−
18
t
+
12
v
=
12
m/
s
Step 3 · Find acceleration (a)
Use a = 6 × c₃ × t + 2 × c₂
a
=
dv/dt:
the
power
rule
again,
on
v
=
3
c₃
t
2
+
2c₂
t
+
c₁
.
Put
t
in.
a
=
6
×
2
×
3
+
2
×
(−9)
d/dt
(
6
t
²
−
18
t
+
12)
=
12
t
−
18
a
=
18
m/
s
2
Check
Put the numbers back in. Both sides should match.
14
=
2
×
3
3
+
(−9)
×
3
2
+
12
×
3
+
5
✓
12
=
3
×
2
×
3
2
+
2
×
(−9)
×
3
+
12
✓
18 = 6 × 2 × 3 + 2 × (−9) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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