SME
University Physics I: Mechanics · Problem type
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.”
Chart
x
=
c₀
+
c₁t
+
c₂
t
2
+
c₃
t
3
x
=
4
+
12
×
3
−
9
×
3
2
+
2
×
3
3
x = 13 m
v
=
dx/dt
=
c₁
+
2c₂t
+
3c₃
t
2
v
=
12
−
2
×
9
×
3
+
3
×
2
×
3
2
v = 12 m/s
a = dv/dt = 2c₂ + 6c₃t
a
=
−18
+
6
×
2
×
3
=
18
m/
s
2
It turns round (v = 0, ringed) at t = 1 s and 2 s.
The tangent to x–t at t has slope v.
This is an example. Type your own values to start a new problem.
x(t)
=
c₃
t³
+
(
c₂
)
t²
+
c₁
t
+
c₀
v(
t (s)
)
=
v (m/s)
a(
t (s)
)
=
a (m/s²)
x
Position
worked out (not typed) · m
Formulas
x
=
c₀
+
c₁
×
t
+
c₂
×
t
2
+
c₃
×
t
3
13
=
4
+
12
×
3
+
(−9)
×
3
2
+
2
×
3
3
v
=
c₁
+
2
×
c₂
×
t
+
3
×
c₃
×
t
2
12
=
12
+
2
×
(−9)
×
3
+
3
×
2
×
3
2
a
=
2
×
c₂
+
6
×
c₃
×
t
18 = 2 × (−9) + 6 × 2 × 3
Assumptions
•
The position is given as x(t) = c₀ + c₁t + c₂t² + c₃t³, x in meters and t in seconds.
•
Power rule: the derivative of tⁿ is ntⁿ⁻¹, so v = dx/dt and a = dv/dt term by term.
•
Where v = 0 and changes sign, the object turns round; a quadratic v(t) can have two such times.
Step-by-step
WE KNOW
c₃ = 2 m/s³, c₂ = −9 m/s², c₁ = 12 m/s, c₀ = 4 m, t = 3 s
FIND
position (x), velocity (v), acceleration (a)
Step 1 · Find position (x)
Use x = c₀ + c₁ × t + c₂ × t² + c₃ × t³
Write
x
(
t
)
with
its
numbers,
put
the
time
in
one
term
at
a
time,
and
add.
x
=
2
t
³
−
9
t
²
+
12
t
+
4
x
=
4
+
12
×
3
+
(−9)
×
3
2
+
2
×
3
3
x
=
4
+
36
−
81
+
54
x
=
13
m
Step 2 · Find velocity (v)
Use v = c₁ + 2 × c₂ × t + 3 × c₃ × t²
v
=
dx/dt:
by
the
power
rule
each
cₙtⁿ
gives
ncₙtⁿ⁻¹,
and
c₀
drops
out.
Put
t
in
and
add.
v
=
6
t
²
−
18
t
+
12
v
=
12
+
2
×
(−9)
×
3
+
3
×
2
×
3
2
v
=
12
−
54
+
54
v
=
12
m/s
Step 3 · Find acceleration (a)
Use a = 2 × c₂ + 6 × c₃ × t
a
=
dv/dt:
the
power
rule
again,
on
v
=
c₁
+
2c₂
t
+
3
c₃
t
2
.
Put
t
in
and
add.
a
=
12
t
−
18
a
=
2
×
(−9)
+
6
×
2
×
3
a
=
−18
+
36
a
=
18
m/
s
2
Check
Put the numbers back in. Both sides should match.
13
=
4
+
12
×
3
+
(−9)
×
3
2
+
2
×
3
3
✓
12
=
12
+
2
×
(−9)
×
3
+
3
×
2
×
3
2
✓
18 = 2 × (−9) + 6 × 2 × 3 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.