Formula‌‌Sheet‌‌-‌‌SPH4U‌ ‌
‌
→
→
→
→
∆dT = ∆d1 + ∆d2 + ∆dn
→
‌∆d = v→∆t
‌∆t =
−
→ →
‌∆d = df inal
→
∆d
v→avg
→
dinitial
‌
sin B
b
‌a = √b2 + c2
− 2bc · cos A
→
∆d = v→f ∆t
−
v→f =
√v→
v→i =
→
→ →
√v − 2a∆d
‌
n Km
h
n 1000m
3600s
=
−
v→f v→i
∆t
a→avg =
∆t =
nm
3.6s
=
‌
−
v→f v→i
a→
− a→∆t
‌‌
F F = μF F N ‌
Fc =
v=
mv 2
r
√
→
∆d =
‌
rF c
m
r=
mv 2
Fc
‌
m=
rF C
v2
‌
1
km
1000
1
3600s
→
2∆d
∆t
− v→
v→f =
→
2∆d
∆t
− v→
a→ =
i
FF
μF
v2
r
‌
‌
v = √ac r ‌
v2
ac
‌
‌
‌sina A =
→
→ 2 ‌
∆d = v→i ∆t + 12 a∆t
‌‌
f
FN =
3.6n Km
s
=
v→i =
v→i =
r=
n
→
2∆d
v→f + v→i
ac =
‌
=
v→f + v→i
2 ∆t
∆t =
‌v→f = v→i + a→∆t ‌
‌v→i = v→f
nm
s
‌‌
∆t =
→
∆d
∆t
−
→
2∆d
∆t2
1→
2 a∆t
−
−v→ ± √v→
i
2
i
2v→i
∆t
→ →
+ 2a∆d
FF
FN
√
‌
∆t =
→
∆d
∆t
v→f ±
Fc
m4π2 r
‌
r=
Fc
m4π2 f 2
‌
m=
Fc
4π2 rf 2
‌
1→ 2
2 a∆t
‌
→ ‌
+ 12 a∆t
−
→
2∆d
2
∆t
→
2vf
∆t
→2
f
→ →
≈ m0 = 0
ac = 4π2 rf 2 ‌
f=
√
r=
ac
4π2 r
ac
4π2 f 2
‌
‌
‌
‌
F
→
net
→
∆d
∆t
→ →
+ 2a∆d
‌
2
−
v→f 2 v→i 2
→
2∆d
‌‌
−
2
2
v→f v→i
→
2a
‌‌
→
→
→
= F1 + F2 + Fn ‌
Fc =
T =
m4π2 r
T2
√
‌
m4π2 r
Fc
r=
T 2F c
m4π2
‌
m=
T 2F c
4π2 r
‌
sinθ
cosθ
initial
‌
v→T x = v→T
× cos θ
v→T y = v→T
× sin θ
θ = tan−1
v→T =
→
√v
→
∆dx =
T =
v→T x
v→T y
( )
2
Tx
+ v→T y
v→i 2
a sin2θ
ac =
‌
− v→
‌∆v→ = v→f inal
‌tanθ =
f
→
‌∆d =
‌
a
2
i
‌a→ =
√v→ − 2a∆d
→
F net = ma→
‌
F c = m4π2 rf 2 ‌
f=
a→ =
‌
a
μF =
v→f =
‌
‌v→ =
4π2 r
T2
√
r=
‌
2
‌
‌
‌
4π2 r
ac
ac T 2
4π2
‌
‌
‌
‌
‌
‌
First‌‌Law:‌‌‌If‌‌the‌‌external‌‌net‌‌force‌‌on‌‌an‌‌object‌‌is‌‌zero,‌‌the‌‌object‌‌will‌‌remain‌‌at‌‌rest‌‌or‌‌continue‌‌to‌‌move‌‌at‌‌a‌‌constant‌‌velocity.‌ ‌
Second‌‌Law:‌‌‌If‌‌the‌‌net‌‌external‌‌force‌‌on‌‌an‌‌object‌‌is‌‌not‌‌zero,‌‌the‌‌object‌‌will‌‌accelerate‌‌in‌‌the‌‌direction‌‌of‌‌the‌‌net‌‌force.‌‌The‌‌magnitude‌‌of‌‌
the‌‌acceleration‌‌is‌‌directly‌‌proportional‌‌to‌‌the‌‌magnitude‌‌of‌‌the‌‌net‌‌force‌‌and‌‌inversely‌‌proportional‌‌to‌‌the‌‌object’s‌‌mass.‌ ‌
Third‌‌Law:‌‌‌For‌‌every‌‌action‌‌force,‌‌there‌‌exists‌‌a‌‌simultaneous‌‌reaction‌‌force‌‌that‌‌is‌‌equal‌‌in‌‌magnitude‌‌but‌‌opposite‌‌in‌‌direction.‌ ‌
‌
‌‌