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unit 2 test 1

what is a scalar

a quantity requiring a number only to describe them

what is a vector

a quantity requiring a number (magnitude) and a direction

how are vectors written diff to scalars

bolded/underlines

is distance a scalar or vector

scalar bc just number

position is what

place relative to referance point (zero)
in metres

vector

what is displacement

change in position
has magnitude and direction

thus vector

final pos minus og pos

what is speed

rate of change of distance
how far travelled per unit of time

scalar

how to find speed

distance travelled/time taken

what is velocity

rate of change of displacement
change in position per unit of time

vector

how to find velocity

displacement/time taken

what does instantaneous speed/velocity show

gives indication if how fast smt is moving at a moment in time

what does average speed/velocity show

an indicatiok of how fast smt is moving over an interval of time

during which the instantaneous speed/velocity has likely changed

thus avg

units for speed/velocity

m/s or km/h
to find each from each you divide or multiply by 3.6 (m/s to km/h is x3.6)

what is acceleration

rate of CHANGE of velocity
change per unit of time

how to find acceleration

triangle velocity/ triangle time
thus its a vector

through motion, objects are either

stationary (zero velocity)
moving w/ constant velocity (zero accel)

or accelerating uniformly/constantly

the 5 variables of the equations of motion are

s= displavement between initial/final position (metres)
u= initial velocity (ms^-1)

v= final velocity (ms^-1)

a=acceleration (ms^-2)

t= time between initial/final pos (s) (only scalar one)

equation of motion 1 (def of accel)

a=triangleV/triangle t
also = v-u/t

or v=u+ut

equations of motion 2 (def of avg velocity)

Vavg= s/t
or Vavg= (u+v)/2

so s/t=(u+v)/2

therefore s=((u+v)t)/2

equations of motion 3 (displacement)

area=s=๐’–๐‘ก+1/2 ๐’‚๐‘ก^2
Since ๐’‚=โˆ†๐’—/๐‘ก so โˆ†๐’—=๐’‚๐‘ก:

๐’”=๐’–๐‘ก+1/2ร—๐‘กร—๐’‚ร—๐‘ก

therefore s=ut+1/2a(t)^2

equations of motion 4

๐’–=๐’—โˆ’๐’‚๐‘ก
thus ๐’”=๐’—๐‘กโˆ’1/2 ๐’‚๐‘ก^2

motions of equations 5

๐‘ก=(๐’—โˆ’๐’–)/๐’‚
therefore ๐’—^2=๐’–^2+2๐’‚๐’”

what is gravity's constant acceleration

9.8ms^-2 downwards

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