Calculus in Year 11–12 Physics
Year 12 Advanced Maths & Physics
Modelling Quantities that Vary: Derivatives & Integrals
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Year 12 Advanced Maths & Physics
Formula Sheet: Key Calculus Links in Physics
Key ideas to review before you start
• Derivatives as rates
v(t) =
ds
,
dt
a(t) =
• Integrals as accumulation If
dv
,
dt
I(t) =
dq
,
dt
E(t) = −N
dΦ
.
dt
dX
= f (t), then
dt
Z
X(t) = f (t) dt + C.
Use a given initial value X(t0 ) = X0 to find C.
• Fundamental Theorem of Calculus (FTC)
Z t=b
dΦ
dt = Φ(b) − Φ(a).
t=a dt
• Work done by a variable force If force depends on position, F (x), then
Z x=b
W =
F (x) dx.
x=a
• Gravitational field varies with distance
g(r) =
GM
,
r2
Z r=b
∆U =
GM m
dr.
r2
r=a
Questions
Q1: Projectile motion with non-constant acceleration (calculus kinematics)
A ball is launched from ground level with initial speed u = 30 m/s at an angle of 40◦ above the
horizontal. Neglect air resistance. The vertical acceleration is not constant and is modelled
by
dvy
= −g − kt, g = 9.8 m/s2 , k = 1.20 m/s3 ,
ay (t) =
dt
for the duration of the flight. Assume horizontal acceleration is zero: ax (t) = 0.
(a) Find vx (t) and vy (t).
(b) Find x(t) and y(t), given x(0) = 0 and y(0) = 0.
(c) Find the time tmax when the projectile reaches maximum height.
(d) Find the maximum height ymax .
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dΦ
Q2: Electromagnetic induction using
(Faraday’s law)
dt
A coil has N = 400 turns. The magnetic flux per turn through the coil is
Φ(t) = 0.020 t2 e−0.50t Wb,
0 ≤ t ≤ 4,
with t in seconds.
(a) Find the induced emf E(t) using E(t) = −N
dΦ
.
dt
(b) Find E(1).
(c) Find the time t > 0 when E(t) = 0.
(d) The coil is connected across a 20 Ω resistor. Find the total charge that flows through the
resistor from t = 0 to t = 4.
Q3: Work and energy with a variable force (integral as area)
A 50 kg sled is pulled along a straight track. The pulling force varies with position x (metres)
as
F (x) = 100 + 20x − 2x2 N,
0 ≤ x ≤ 10.
Assume the sled starts from rest and friction is negligible.
(a) Find the work done by the force from x = 0 to x = 10.
(b) Find the speed of the sled at x = 10.
Q4: Gravitation: moving between orbits (field varies with r)
A spacecraft of mass m = 1200 kg moves away from Earth from radius r = RE to r = 2RE
from Earth’s centre (so r is distance from Earth’s centre). Use
GM = 3.99 × 1014 m3 /s2 ,
RE = 6.37 × 106 m.
(a) Using an integral, find the increase in gravitational potential energy ∆U for the spacecraft
from RE to 2RE .
(b) If the spacecraft is launched from r = RE with speed v0 and arrives at r = 2RE with
zero speed, find v0 .
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Solutions
Q1: Given: u = 30, θ = 40◦ , ax (t) = 0, ay (t) = −g − kt, g = 9.8, k = 1.20.
(a) Velocity components
Initial components:
ux = u cos θ
uy = u sin θ
ux = 30 cos 40◦
uy = 30 sin 40◦
Horizontal:
ax (t) =
dvx
=0
Zdt
vx (t) =
0 dt + C
vx (t) = C
Use vx (0) = ux :
vx (0) = C
ux = C
vx (t) = u cos 40◦
Vertical:
dvy
= −g − kt
dt
Z
vy (t) =
(−g − kt) dt
k
vy (t) = −gt − t2 + C
2
Use vy (0) = uy :
uy = C
k
vy (t) = u sin 40◦ − gt − t2
2
(b) Position components
Horizontal:
dx
= vx (t) = u cos 40◦
dt
Z
x(t) =
u cos 40◦ dt
x(t) = (u cos 40◦ )t + C
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Use x(0) = 0:
0=C
x(t) = (u cos 40◦ )t
Vertical:
dy
k
= vy (t) = u sin 40◦ − gt − t2
dt
2
Z k
u sin 40◦ − gt − t2 dt
y(t) =
2
g
k
y(t) = (u sin 40◦ )t − t2 − t3 + C
2
6
Use y(0) = 0:
0=C
g
k
y(t) = (u sin 40◦ )t − t2 − t3
2
6
(c) Time of maximum height
Maximum height occurs when vy (t) = 0.
k
0 = u sin 40◦ − gt − t2
2
k 2
0 = − t − gt + u sin 40◦
2
k 2
0 = t + gt − u sin 40◦
2
Quadratic in t:
t=
−g ±
p
g 2 + 2k(u sin 40◦ )
k
Take the positive root:
tmax =
−g +
p
g 2 + 2k(u sin 40◦ )
k
(d) Maximum height
ymax = y(tmax )
g
k
ymax = (u sin 40◦ )tmax − t2max − t3max
2
6
g
k
ymax = (u sin 40◦ )tmax − t2max − t3max
2
6
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y
u
40◦
(0, 0)
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x
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Year 12 Advanced Maths & Physics
Q2: Given: N = 400, Φ(t) = 0.020 t2 e−0.50t .
(a) Induced emf
dΦ
dt
Φ(t) = 0.020 t2 e−0.50t
E(t) = −N
Differentiate using product rule on t2 · e−0.50t :
dΦ
d 2
−0.50t
2 d
−0.50t
= 0.020
(t ) · e
+ t · (e
)
dt
dt
dt
= 0.020 2t e−0.50t + t2 (−0.50)e−0.50t
= 0.020 e−0.50t 2t − 0.50t2
So
E(t) = −400 0.020 e−0.50t (2t − 0.50t2 )
E(t) = −8 e−0.50t (2t − 0.50t2 )
1 2
−0.50t
E(t) = −8 e
2t − t
2
E(t) = −8e
−0.50t
1
2t − t2
2
V
(b) Find E(1)
1 2
E(1) = −8e
2(1) − (1)
2
1
E(1) = −8e−0.50 2 −
2
3
E(1) = −8e−0.50
2
−0.50
E(1) = −12e−0.50 V
(c) Solve E(t) = 0 for t > 0
1
0 = −8e−0.50t 2t − t2
2
Since e−0.50t ̸= 0, we solve:
1
0 = 2t − t2
2 1
0=t 2− t
2
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t=0
or
t=0
or
1
2− t=0
2
t=4
For t > 0:
t= 4s
(d) Total charge through a 20 Ω resistor (0 to 4 s)
E(t)
R
Z 4
I(t) dt
qtotal =
0
Z 4
E(t)
qtotal =
dt
R
0
Z 4
−N dΦ
qtotal =
dt
R dt
0
I(t) =
Use FTC:
qtotal =
−N
[Φ(4) − Φ(0)]
R
Compute Φ(4) and Φ(0):
Φ(4) = 0.020 · 42 e−0.50·4
Φ(4) = 0.020 · 16 e−2
Φ(4) = 0.32e−2
Φ(0) = 0.020 · 02 e0
Φ(0) = 0
So
−400
0.32e−2 − 0
20
qtotal = −20 0.32e−2
qtotal =
qtotal = −6.4e−2
Magnitude of charge that flows:
|qtotal | = 6.4e−2 C
Φ(t) varies
Coil
R
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Q3: Given: m = 50, F (x) = 100 + 20x − 2x2 , 0 ≤ x ≤ 10.
(a) Work done
Z 10
F (x) dx
W =
0
Z 10
W =
100 + 20x − 2x2 dx
0
10
x2
x3
W = 100x + 20 ·
−2·
2
3 0
10
2
W = 100x + 10x2 − x3
3
0
2 3
2
W = 100(10) + 10(10 ) − (10 ) − 0
3
2000
W = 1000 + 1000 −
3
2000
W = 2000 −
3
4000
W =
3
W =
4000
J
3
(b) Speed at x = 10
By work-energy:
W = ∆K
1
W = mv 2 − 0
2
4000
1
= (50)v 2
3
2
4000
= 25v 2
3
4000
v2 =
75
160
2
v =
r3
160
v=
3
r
v=
160
m/s
3
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F (x) (N)
200
150
100
50
F (x) = 100 + 20x − 2x2
x (m)
1
2
3
4
5
6
7
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8
9
10
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Year 12 Advanced Maths & Physics
Q4: Given: m = 1200, GM = 3.99 × 1014 , RE = 6.37 × 106 .
(a) Increase in gravitational potential energy
Gravitational force magnitude:
F (r) =
GM m
r2
The increase in gravitational potential energy from RE to 2RE is the work done against
gravity:
Z 2RE
GM m
dr
r2
RE
Z 2RE
∆U = GM m
r−2 dr
∆U =
RE
2R
∆U = GM m −r−1 R E
E
1
1
+
∆U = GM m −
2RE
RE
1
∆U = GM m
2RE
GM m
∆U =
2RE
∆U =
GM m
2RE
Substitute numbers:
(3.99 × 1014 )(1200)
2(6.37 × 106 )
4.788 × 1017
∆U =
1.274 × 107
∆U ≈ 3.76 × 1010
∆U =
∆U ≈ 3.76 × 1010 J
(b) Launch speed to arrive at 2RE with zero speed
Use energy conservation:
Ki − Kf = ∆U
1
mv 2 − 0 = ∆U
2 0
2∆U
v02 =
m
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Using ∆U =
Year 12 Advanced Maths & Physics
GM m
:
2RE
2 GM m
·
m 2RE
GM
v02 =
R
rE
GM
v0 =
RE
v02 =
r
v0 =
GM
RE
Numerically:
r
v0 =
3.99 × 1014
6.37 × 106
v0 =
p
6.26 × 107
v0 ≈ 7.91 × 103
v0 ≈ 7.91 × 103 m/s
2RE
Earth
RE
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