thesuvat equations$ - wordpress.com · thesuvat equations$ pg3638 # # terminology$ •...
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The SUVAT Equations Pg 36-‐38 ©
cgraham
physics.com 2015
Terminology • s= displacement • u = ini?al / star?ng speed • v = final speed / velocity • a = accelera?on • t = ?me
© cgraham
physics.com 2015
Deriving the equations from v – t graph
• Accelera?on = gradient • 𝑎= 𝑣−𝑢/𝑡 • 𝑎𝑡=𝑣−𝑢 • 𝑣=𝑢+𝑎𝑡 àvelocity equa?on
• Area under v-‐t graph = distance
velocity
?me t
v
u
𝐴= 1/2 (𝑣−𝑢)𝑡
𝐴=𝑢×𝑡
S = total area = = 1/2 (𝑣−𝑢)𝑡+𝑢𝑡= 1/2 (𝑣−𝑢)𝑡/𝑡 ×𝑡+𝑢𝑡 𝑠=𝑢𝑡+ 1/2 𝑎𝑡↑2 à distance equa,on
a
© cgraham
physics.com 2015
We know… • 𝑎= 𝑣−𝑢/𝑡 • 𝑡= 𝑣−𝑢/𝑎 • s =𝑢𝑡+ 1/2 𝑎𝑡↑2 =𝑢( 𝑣−𝑢/𝑎 )+ 1/2 𝑎(𝑣−𝑢/𝑎 )↑2
• 𝑠𝑎=𝑢(𝑣−𝑢)+ 1/2 (𝑣−𝑢)↑2 • 2sa= 𝑣↑2 − 𝑢↑2 • 𝑣↑2 = 𝑢↑2 +2𝑠𝑎 à ?meless EQ ©
cgraham
physics.com 2015
Average displacement • Average speed = 𝑢+𝑣/2 • 𝑣↓𝑎𝑣 = 𝑠/𝑡 • 𝑠=(𝑢+𝑣/2 )𝑡 • à average displacement
© cgraham
physics.com 2015
v = u + at
s = ut + at 2 1 2
v2 = u2 + 2as
© cgraham
physics.com 2015
Solving problems using equations of motion for uniform acceleration
Topic 2: Mechanics 2.1 – Motion
EXAMPLE: How far will Pinky and the Brain go in 30.0 seconds if their acceleration is 20.0 m s -2? a = 20 m/s2
KNOWN Given
u = 0 m/s Implicit t = 30 s Given
WANTED s = ?
FORMULAS
v = u + at v2 = u2 + 2as
•t is known - drop the timeless eq’n. •Since v is not wanted, drop the velocity eq'n:
SOLUTION
s = 0(30) + 20(30)2 1 2
s = 9000 m
s = ut + at2 1 2
s = ut + at2 1 2
© cgraham
physics.com 2015
Solving problems using equations of motion for uniform acceleration
Topic 2: Mechanics 2.1 – Motion
EXAMPLE: How fast will Pinky and the Brain be going at this instant?
a = 20 m/s2 KNOWN
Given
u = 0 m/s Implicit t = 30 s Given
WANTED v = ?
FORMULAS
v = u + at v2 = u2 + 2as
•t is known - drop the timeless eq’n. •Since v is wanted, drop the displacement eq'n:
SOLUTION
s = ut + at2 1 2
v = u + at v = 0 + 20(30) v = 600 m s-1
© cgraham
physics.com 2015
Solving problems using equations of motion for uniform acceleration
Topic 2: Mechanics 2.1 – Motion
EXAMPLE: How fast will Pinky and the Brain be going when they have traveled a total of 18000 m? a = 20 m/s2
KNOWN Given
u = 0 m/s Implicit s = 18000 m Given
WANTED v = ?
FORMULAS
v = u + at v2 = u2 + 2as
•Since t is not known - drop the two eq’ns which have time in them.
SOLUTION
s = ut + at2 1 2
v2 = u2 + 2as v2 = 02 + 2(20)(18000) v = 850 m s-1
© cgraham
physics.com 2015
Example • A cyclist slows uniformly from a speed of 7.5m𝑠↑−1 to a speed of 2.5m𝑠↑−1 in a ?me of 5.0s.
• Calculate a) the accelera?on b) the distance moved in 5.0s
Solu?on • 𝑣=𝑢+𝑎𝑡 • 𝑎= 𝑣−𝑢/𝑡 = 2.5−7.5/5 = −5/5 �=−1.0𝑚𝑠↑−2 nega?ve sign = decelera?on
b) 𝑣↑2 = 𝑢↑2 +2𝑠𝑎 𝑠= 𝑣↑2 − 𝑢↑2 /2𝑎 = 2.5↑2 − 7.5↑2 /2(−1.0) =25𝑚
© cgraham
physics.com 2015
Example • A driver of a car travelling at 25 m 𝑠↑−1 along a road applies the brakes. The car comes to a stop in 150m with a uniform decelera?on.
• Calculate a) the ?me the car takes to stop b) the decelera?on of the car
Solu?on • 𝑠= 𝑢+𝑣/2 𝑡 𝑡= 2𝑠/𝑢+𝑣
• 𝑡= 2×150/25+0 =12𝑠
• b) v = u + at • 𝑎= 𝑣−𝑢/𝑡 = 0−25/12 �=−2.1 m𝑠↑−2
© cgraham
physics.com 2015
Example • A car starts from rest and reaches a speed of 36 m 𝑠↑−1 in 12s. What is the accelera?on of the car?
Solu?on • 𝑎= 𝑣−𝑢/𝑡 = 36−0/12 =3𝑚𝑠↑−2
• The car then brakes and stops with a decelera?on of 5.0 𝑚𝑠↑−2 . How far does the car travel before stopping?
Solu?on • 𝑣↑2 = 𝑢↑2 +2𝑠𝑎 • 𝑠= 𝑣↑2 − 𝑢↑2 /2𝑎 = 0− 36↑2 /2(−5) =129.6𝑚 ~130𝑚
© cgraham
physics.com 2015
Example • A car is travelling with constant speed of 80𝑘𝑚ℎ↑−1 in a school zone. A police car starts at rest as the speeder passes it and accelerates at a constant rate of 8.0 𝑘𝑚ℎ↑−2 .
• When does the police car catch the speeding car?
• 𝑠↓𝑠 = 𝑠↓𝑝 at ?me t • 𝑣↓𝑠 = 80 𝑘𝑚ℎ↑−1 =22.2m 𝑠↑−1
• 𝑠↓𝑠 = 𝑣↓𝑠 t • 𝑠↓𝑝 =0+1/2 𝑎𝑡↑2 • 𝑣↓𝑃 = 8.0 𝑘𝑚ℎ↑−2 =2.22m 𝑠↑−2
distance
?me t
d
𝑣𝑡= 1/2 𝑎𝑡↑2
𝑡= 22.22/1.11 =20.0𝑠
© cgraham
physics.com 2015