Derivations

Question 1

Derive v = u + at starting with a = dv/dt

Answer 1

integrate with respect to time
ds/dt = at + k
at t = 0 ds/dt = u so k = 0
at t = t  ds/dt = v
thus v = u + at

Question 2

Derive s = ut + 1/2at^2 starting at ds/dt = v = u + at

Answer 2

integrate with respect to time
s = ut + 1/2at^2 + c
at t = 0, s = 0 so c = 0
s = ut + 1/2at^2

Question 3

Derive v^2 = u^2 + 2as starting at v = u + at

Answer 3

Square both sides
v^2 = (u+at)(u+at)
v^2 = u^2 + 2uat + a^2t^2
v^2 = u^2 + 2a(ut + 1/2at^2)
s = ut + 1/2at^2
v^2 = 2as

Question 4

Conical Pendulum

\
|θ \ T
|___\
Tsinθ

Answer 4

Tcosθ = mg      divide the two
Tsinθ = F        divide the two

tanθ = F/mg

F = Centripetal force
mg = W

Question 5

Derive escape velocity

Answer 5

Total energy on planet’s surface = total energy at infinity = 0
Ek + Ep = 0
1/2mv^2 + ( - GMm/r ) = 0 where M is the mass of the planet and r it’s radius
v^2 = 2GM/r
so v = √2GM/r

Question 6

derive r = mv/qB

Answer 6

F = mv^2/r      and      F = Bqv
so
mv^2/r = qvB
thus
r = mv/qB

Question 7

derive a = -wy^2 starting at y = Asinwt

Answer 7

y = Asinwt
dy/dt = (Asinwt)
v = Awcoswt
dv/dt = (Awcoswt)
a = -Aw^2sinwt
since Asinwt = y
then
a = -w^2y

Question 8

derive a = -wy^2 starting at y = Acoswt

Answer 8

y = Acoswt
dy/dt = (Acoswt)
v = - Awsinwt
dv/dt = (Awsinwt)
a = -Aw^2coswt
since Acoswt = y
then
a = -w^2y

Question 9

derive v = w√A^2 - y^2 starting at v = Awcoswt

Answer 9

v = Awcoswt
1 - sin^2wt = cos^2wt
√1-sin^2twt = coswt
v = ±Aw√1-sin^2wt
y = Asinwt
y^2 = A^2sin^2wt
v = ±Aw√1 - y^2/A^2
v = ±√A^2 w√1 - y^2/A^2
v = ±w√A^2( 1 - y^2/A^2)
v = ±w√A^2 - y^2

Question 10

Derive Kinetic Energy for SHM

Answer 10

Ek = 1/2mv^2
since v = ±w√A^2 - y^2
Ek = 1/2 m (v = ±w√A^2 - y^2)^2
so
Ek = 1/2mw^2 (A^2 - y^2)

Question 11

Derivation of d = λ/4n

for glass lenses with a coating such as magnesium flouride

Answer 11

there is a phase change of πwhen both rays are reflected
the coating has a thickness of d
optical path difference = 1/2λ for destructive interference
optical path coating = 2nd
thus
1/2λ = 2nd
d = λ/4n

Question 12

Derivation of n=taniₚ

Answer 12

n=siniₚ/sinr
but iₚ +r = 90° 
so n = siniₚ/sin(90-iₚ)
thus n=siniₚ/cosiₚ
so n =taniₚ