Complete Guide to Motion: Distance, Velocity, Acceleration & Projectile Physics

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videos in this video I've covered all the outlines in the syllabus as shown in the

figure distance and displacement distance is the total length of the path traveled by an object

distance is a scalar quantity and its unit is the meter displacement is the directed distance from the start to the

end points displacement is a vector quantity and its unit is the meter when a ball travels along a circular track

from point A to point B the distance traveled from a to B is half of the circumference of the circle the

displacement is the diameter of a circle and its direction point from A to B or downward or South when a ball travels

back to point a the distance traveled from point A to return at Point a is equal to the circumference of a circle

the displacement is equal to zero this is because the ball has returned to its original

position speed and velocity speed is the rate of change in distance speed is a scalar quantity and its unit is

m/s velocity is the rate of change in displacement velocity is a vector quantity and its unit is

m/s the sign positive or negative in front of the Velocity indicates the direction of the object's motion the

magnitude of velocity is equal to speed when an object travels in a straight line and does not change its

direction for example a car travels with constant speed up and down the hill as shown while its speed remains constant

its velocity is not constant due to the changing direction of movement a car is traveling at an initial velocity U for T

seconds reaches a final velocity V and covers a displacement of s m assuming constant

acceleration the average velocity is U + V / 2 the total displacement moved is s m the total time is T seconds the

equation for displacement can be rearranged as s is U + V / 2 multiplying of T this formula is used for only

constant accelerated Motion in a straight line

acceleration acceleration is the rate of change in velocity acceleration is a vector

quantity and its unit is m/s squared therefore the equation of acceleration is AAL vus U / t

this formula comes from IG csse this equation can be rearranged as v = u + a t this formula is used for

only constant accelerated Motion in a straight line when V equals u meaning that the acceleration is zero there is

no change in velocity and the object moves with constant velocity when V is greater than u

meaning that the acceleration is positive the velocity is increasing and the object is accelerating

when V is less than u meaning that the acceleration is negative the velocity is decreasing and the object is

decelerating motion graphs displacement time graph its gradient is change in displacement over change in time which

is equal to Velocity so the gradient of a displacement time graph is velocity

while its area under graph does not have a meaning velocity time graph its gradient is change in velocity over

change in time which is equal to acceleration so the gradient a velocity time graph is

acceleration the area under graph in this case is a trapezium shape which is U + V over 2 multiplying of T so its

area under graph is displacement moved the graph shows how the displacement of an object moving in a

straight line varies over time t as shown recall that the gradient of a displacement time is

velocity as the gradient increases the velocity increases as the gradient is constant

the velocity is constant as the gradient decreases the velocity decreases as the gradient is zero the

velocity is zero as the gradient is negative and increasing the velocity is increasing in the opposite direction as

the gradient is negative and constant the velocity is constant in the opposite direction as the gradient is negative

and decreasing the velocity is decreasing in the opposite direction sketch the velocity time graph of the

object over the same time interval as the previous displacement time graph is shown recall that the gradient of a

velocity time graph is acceleration at this section we draw the graph like this to represent increasing

velocity at a constant rate so the gradient is positive and constant indicating constant

acceleration at this section we draw the graph like this to represent constant velocity so the gradient is zero

indicating no acceleration at this section we draw the graph like this to represent decreasing

velocity at a constant rate so the gradient is negative and constant indicating constant

deceleration at this section we draw the graph like this to represent zero velocity so the gradient is zero and no

acceleration at this point the velocity will change its sign indicating that the motion's

direction will change at this section we draw the graph like this to represent increasing

velocity in the opposite direction at a constant rate so the gradient is negative and constant indicating the

acceleration is negative and constant this shows that the velocity and acceleration are in the same direction

at this section we draw the graph like this to represent constant velocity in the opposite direction

so the gradient is zero indicating no acceleration at this section we draw the graph like this to represent decreasing

velocity in the opposite direction at a constant rate so the gradient is positive and constant indicating the

acceleration is positive and constant this shows that the velocity and acceleration are in opposite

directions the area under graph above x-axis is the displacement moved to the right or upward the area under graph

below x-axis is the displacement moved to the left or downward sketch the acceleration time graph of the object

over the same time interval as the previous velocity time graph is shown at this section we draw the graph like this

to represent constant positive acceleration at this section we draw the graph like this to represent zero

acceleration at this section we draw the graph like this to represent constant negative

acceleration at this section we draw the graph like this to represent Z acceleration at this section we draw the

graph like this to represent constant negative acceleration at this section we draw the

graph like this to represent zero acceleration at this section we draw the graph like this to represent constant

positive acceleration exam style question question one the variation with time T

of the Velocity V of the car is shown below what is the average velocity of the toy car for the journey shown by the

graph average velocity is calculated by dividing total displacement by total time the total displacement is equal to

the net area under the velocity time graph the area under graph above xaxis is positive which is equal to 3 * 10

resulting in 30 m the area under graph below x-axis is negative which is equal to 6 * 5

resulting in -30 M since the total displacement is 30 - 30 resulting in 0o therefore the average velocity is 0 / 15

resulting in 0 m/s question two a mass on the end of a spring bounces up and down as shown

after being released at time T equals 0 sketch the graph of velocity varies with time recall that the gradient of the

displacement time graph is velocity at this point the gradient is zero causing the veloc velocity is also zero in this

section the positive gradient is increasing causing the velocity to be positive and also

increase at this point the positive gradient is maximum causing the velocity to be maximum positive in this section

the positive gradient is decreasing causing the positive velocity is also decreasing at this point the gradient is

zero causing the velocity to be also zero in this section the gradient becomes increasing negative causing the

velocity to become more negative at this point the gradient is maximum negative causing the velocity is also negative

maximum in this section the negative gradient is decreasing causing the negative velocity to be less negative

value at this point the gradient is zero causing the velocity is also zero question three the graph shows how

the velocity V of an object moving in a straight line varies over time t = 0 to tal T sketch the displacement s of the

object in the time tal 0 to t equal T recall that the gradient of displacement time graph is velocity at this point the

velocity is zero so the gradient of displacement time graph to be zero in this section the positive velocity is

increasing so the positive gradient of displacement time graph to increase at this point the positive

velocity is maximum so the positive gradient of displacement time graph is maximum in this section the positive

velocity is decreasing so the positive gradient of displacement time graph to decrease at this point the velocity is

zero so the gradient of displacement time graph to be zero in this section the velocity becomes increasingly

negative so the gradient of displacement time graph to be negative and increasing at this point the negative

velocity is maximum so the negative gradient of displacement time graph is maximum in this section the negative

velocity is becoming less negative so the negative gradient of displacement time graph to

decrease at this point the velocity is zero so the gradient of displacement time graph is also zero question four

the graph shows how the acceleration of an object moving in a straight line varies with time the object starts from

rest sketch the graph of the variation with time of the velocity of the object over the same time interval recall that

the gradient of velocity time graph is acceleration at this point the acceleration is zero so the gradient of

velocity time graph is zero in this section the positive acceleration is increasing so the gradient of velocity

time graph is positive to increase at this point the positive acceleration is maximum so the positive

gradient of velocity time graph is maximum in this section the positive acceleration is decreasing so the

positive gradient of velocity time graph decreases at this point the acceleration is zero so the gradient of velocity time

graph is zero the equations of motion or kinematic equations are often called

suat equations suat stands for displacement initial velocity final velocity

acceleration and time these equations are used to describe motion with constant acceleration in a straight line

they involve five physical quantities for Vector quantities displacement s initial velocity U final velocity a and

acceleration a and one scalar quantity time T we got the first equation from the definition of average velocity like

we talked about before we got the second equation from the definition of acceleration like we talked about before

we substituted the second equation into the first equation like this and then we rearranged the equation to get the third

equation like this we got the fourth equation by combining the first and second

equations we rearranged the second equation to make time T the main variable like this then we put this

value for time T into the first equation like this we rearranged the equation to get the fourth equation like this when

using kinematic equations we place a negative sign in front of vector quantities that represent present

direction to the left or downward exam style questions question one a bicycle Brak so

that it undergoes uniform deceleration from a speed of 8 m/s to 6 m/s over a distance of 7 m if the deceleration of

the bicycle remains constant what further distance will it travel before coming to rest first let's calculate the

bicycle's acceleration since all Vector quantities in this case are directed to the right

will assign a positive sign to them when the bicycle travels 7 m its displacement s is 7 m its initial velocity U is 8 m/s

its final velocity V is 6 m/s and we need to find its acceleration a we can use the equation V ^2 = U ^2 + 2 a s

let's rearrange this equation to make acceleration a the main variable like this we substitute v = 6 s = 7 and U = 8

we get a = -2 m/ second squared the negative sign indicates deceleration meaning the bicycle is slowing down to

the left now let's calculate how far the bicycle travels before coming to a stop from an initial velocity of 6

m/s we need to find the displacement s we know the initial velocity U is 6 m/s the final velocity RV is zero and the

acceleration is -2 m/s squared due to deceleration we can use the equation V ^2 = U ^2 + 2 a s again let's rearrange

this equation to make displacement s the main variable like this we substitute V = 0 U = 6 and a = -2 we get S = 9 m

question two a moving body undergoes uniform acceleration while traveling in a straight line between points x y and

Zed the distance XY and Y Z are both 40 m the time to travel from X to Y is 12 seconds and from y to Z is 6

seconds what is the acceleration of the body all Vector quantities in this question are directed to the right will

assign a positive sign to them between points X and Y its displacement s is 40 m its initial velocity is U its

acceleration is a and its time T is 12 seconds we can use the equation s equal u t + half of a t

^2 we substitute s = 40 and T = 12 and then we simplify the equation like this between points X and Zed its

displacement s is 80 M its initial velocity is U its acceleration is a and its time T is 18 seconds we can use the

equation s equals U t+ half of a t ^2 again we substitute S = A and T = 18 and then we simplify the equation

like this we can solve these two equations together to find the acceleration a by eliminating of U like

this we get a = 0.37 m/s squared to two significant figures Free Fall is the motion has only

one force acting upon it which is the gravitational force weight without air resistance causing it travel with

constant acceleration as 9.81 m/s squared we can use the equations for motion with constant acceleration in a

straight line to describe Free Fall remember to put a negative sign in front of any quantities like displacement

initial velocity final velocity an acceleration that go downward for example a boy throws a

stone vertically up into the air with a velocity of 6 m per second the stone reaches a maximum height and falls into

the sea which is 12 M below the point of release as shown in a figure a calculate the velocity and time at which the stone

hits the water surface we set the upward direction is positive and downward direction is negative when the boy

throws a stone to reach the sea its displac m s = -12 M negative indicating that it's downward its initial velocity

U = 6 m/ second its final velocity equals V its acceleration a equal 9.81 m/s squared negative indicating that

it's downward and its time equals T we use the equation V ^2 = U ^2 + 2 a s to find the final velocity r v We

substitute U equal 6 A = 9.81 and S = -12 we get V = -16 m per second for two significant figures the negative sign

indicates that it's downward now let's use the equation s equal u t + half of a t ^2 to find the

time T we substitute s = -12 U = 6 and a = 9.81 we rearrange and simplify the

equation like this and then we solve the equation for time T like this we reject the negative answer for time so time T

equals 2.3 for two significant figures B calculate the highest height at which the stone can be reached from the water

surface at the highest point the Stone's velocity is zero when the stone is thrown up to its highest point its

displacement equals s its initial velocity U U = 6 m/s its final velocity V equal 0 m/s and its acceleration a

equal 9.81 m/s sared negative indicating that it's downward we use the equation v² = U ^2 + 2 a s let's rearrange this

equation to make displacement s the main variable like this we substitute V = 0 U = 6 and a =

9.81 we get height s from the start Point equals 1.83 5 m so the highest height from C equals 14 M for two

significant figures C sketch the graphs of the Stone's motion until it hits the C

displacement time velocity time and acceleration time we will give going up is positive and going down is negative

at the start the displacement s is zero when the stone is thrown up its displacement s increases until it

reaches 1.8 m upwards so we draw the displacement time graph like this the graph is a curve

that decreas in gradient because its velocity is decreasing from this point the Stone

Falls back to the ca so its displacement decreases until it's back where it's started then it goes down to -12 M when

it hits the C so we draw the displacement time graph like this the graph is a curve that increasingly

negative gradient because its velocity is becoming more negative at the start the velocity V is 6 m/s upwards when the

stone is thrown up its velocity V decreases from postive 6 to its 0 m/s at the highest point so we draw the

velocity time graph like this the graph is a straight line that constant negatively gradient because its velocity

is decreas inreasing at constant rate from this point the Stone Falls Back to the Sea so its velocity changes

Direction and increases from zero to six until it's back where it started then it goes down to -16 m/s when it hits the C

so we draw the velocity time graph like this the graph is a straight line that constant negatively gradient because its

velocity is becoming more negative at constant rate the stone is falling freely which means it's only affected by

gravity gravity pulls it down at a steady rate of 9.81 m/s squared so we draw the acceleration time

graph like this this horizontal line shows that the acceleration is constant at 9.81 m per second squared and

downward the experiment to determine the acceleration of free fall we set up the experiment as shown the independent

variable of this experiment is the height H that a ball bearing to fall the dependent variable is the time taken T

to fall the control variables are the same mass same diameter or same size of the ball birthing the methods to measure

the measurements as follows to measure the height H from the bottom of a ball bearing to the surface of trapo using a

rule to turn on the electromagnet and a ball bearing Falls towards the trap door the timer starts to measure the time t

as the electromagnet turns on and stopping as a ball bearing reaches the trapo repeat the reading of time T at

same height H for three times and find the average repeat steps 1 to four for six different heights as

0.10 0.15 0.20

0.25 0.30 and 0.35 M record the results in the table plot

the graph of time t^2 on your axis and the height H at on x-axis as shown the method of analysis of this experiment as

follows when a ball bearing Falls the displacement s is netive H indicating downward the initial velocity U is zero

the final velocity a ve is unknown the acceleration is negative G the acceleration due to gravity negative

indicating downward and the time is T we can use the equation s equals u t + half of a t

^2 we substitute s = h u = 0 and a = NE G we rearrange and simplify the equation like this from the graph its gradient is

change in y / change in X which is T ^2 / H we rearrange the equation to get H over T ^2 which is 1 /

gradient we substitute this equation into this equation we get G is 2 / by gradient exam style

questions a student takes measurement to determine a value for the acceleration of Free Fall some of the apparatus used

is Illustrated in figure below the student measure the vertical distance D between the base of the electromagnet

and the bench the time t for an iron ball to fall from the electromagnet to the bench is also measured corresponding

values of T ^2 and D are shown in figure a on the graph draw the line of best fit for the points so we draw the graph like

this B State and explain why there is a non-zero intercept on the graph the distance D is the distance the ball

Falls plus the diameter of the ball because we didn't measure from the bottom of the ball see determine the

student value for one the diameter of ball we can read the Y intercept on the graph which is 1 1.5

cm two the acceleration of free fall when a ball Falls the displacement s is negative D indicating downward the

initial velocity U is zero the final velocity is unknown the acceleration is netive G acceleration of Free Fall

negative indicating downward and the time is T we can use the equation s equal u t + half of a

t^2 we substit itute s = d u = 0 and a = g we rearrange and simplify the equation to make G the main variable like this

from the graph its gradient is change in y / change in X which is D / T ^2 so this is the

gradient we get G equal 2 * gradient we find the change in y on the graph is

0.475 M and the change in x is 0.10 seconds squared so we get the gradient is for.

75 m/s squared and then we substitute gradient into this equation so G equal 9.5 m/s

squared the projectile motion the projectile motion is a two-dimensional motion consisting of a horizontal

component with uniform velocity and a vertical component with uniform acceleration due to gravity this occurs

when a constant downward Force such as gravity acts on an object without air resistance the vertical component of the

motion has uniform acceleration G so we can use kinematic equations the horizontal component of

the motion has no acceleration so the velocity stays the same and we can only use one equation shown here the time T

is the same for both components of the motion let's compare free fall and projectile motion when a green ball

grown horizontally would speed you and a blue ball dropped from the same height at the same time like as shown at the

start both balls have zero initial Vertical Velocity every second both balls fall the same displacement and

have the same downward velocity so they take the same time to fall and have the same acceleration due

to gravity so both balls will reach the ground at the same time and with the same acceleration due to gravity but

they will hit the ground at different speeds this is because the blue Ball's hitting

speed is just v y downward like this and the green Ball's hitting speed is the combination of v y and you using the

Pythagorean theorem like this and its direction is at an angle Theta to the horizontal like this so the blue Ball's

motion is free fall no horizontal component and a constant acceleration G in vertical component the kinematic

equations can use for this the green Ball's motion is the projectile motion no acceleration and constant velocity in

horizontal component and constant acceleration G in vertical component when a ball thrown from the ground with

a speed U at angle of elevation of theta where air resistance is negligible the ball travels along the

curved projectile path as shown at the start we can break the initial velocity U into a horizontal component U cos

Theta and a vertical component U sin Theta as the ball goes up the vertical

velocity decreases but the horizontal velocity stays the same as the ball reach the highest point the vertical

velocity is zero but the horizontal velocity stays the same as the ball falls back down the vertical velocity

increases downward but the horizontal velocity stays the same when the ball reaches the ground the vertical velocity

is the same as it was at the start but opposite direction and the horizontally velocity stays the same at time T the

ball traveled a distance s x horizontally and a distance s y upward we can break down its motion into

horizontal component and vertical component like shown in the table the time T is a common factor for both the

horizontal and vertical components of motion the speed V at this time can be found by combining the horizontal

velocity and the vertical velocity using the Pythagorean theorem like this the D direction of the Velocity V at this time

is the angle beta to the horizontal like this let's calculate the range R in term of the gravity G the angle Theta and the

initial velocity u in the horizontal component we get our equals U CU Theta * T let's give this the first equation the

time T is the same for both components so we need to find T in the vertical component and then put into the first

equation in the vertical component the displacement s is zero the initial velocity U is U sin Theta upward the

acceleration a is netive G downward and the time equals T we use the equation s equals u t + half of a

t^2 we substitute S = 0 u = u sin Theta and a equals g we rearrange and simplify the equation like this and let's call

this the second equation we put the second equation into the first equation like this we use the fact that 2 cos

Theta sin Theta equal sin 2 Theta so we get the range R is U ^ 2 sin 2 Theta / G the equation for the range R is shown

that the range R is longest When Gravity G and the initial speed U are constant which happens when sin 2 Theta is 1 this

means 2 Theta is 90° so Theta is 45° so the longest distance R is achieved when when the angle of

projection is 45° with a constant initial speed U like shown in the diagram that the red path shows the

motion of the object without air resistance this is called projectile motion the Green Path shows the motion

of the object with a resistance this causes the maximum height and the range are to be

shorter exam style questions a ball is thrown from A to B as shown in figure the ball is thrown

with an initial velocity V at 60° to the horizontal the variation with time T of the vertical component of the velocity

of the ball from tal 0 to T = 0.60 seconds is shown in the graph assume air resistance is

negligible one complete the graph for the time initial the ball reaches B the straight line graph is shown which shows

the vertical velocity of the ball the gradient of the Velocity time graph represents

acceleration this shows that the acceleration is a negative constant which Remains the Same for the vertical

component motion of the projectile motion so we draw the graph like this two calculate the maximum height reached

by the ball the area under the graph of velocity time is displacement

moved the vertical velocity is zero at the maximum height so the distance moved equals area under graph between tal 0 to

T = 0.6 seconds this is because the vertical velocity is zero at T = 0.6

seconds so we get the distance equals half of 5.9 * 0.6 = 1.77 m and the maximum height is 1.8 m for two

significant figures three calculate the horizontal component of the velocity of the ball at time tal 0 second we break

down the initial velocity a V at t equal 0 seconds the initial horizontal velocity is v t cos 60 and the initial

Vertical Velocity is V sin 60 in the graph the initial Vertical Velocity is 5.9 m/s

so V sin 60 = 5.9 and we solve for V like this we substitute V in the initial horizontal velocity like this so we get

the initial horizontal velocity is 3. for m/s for two significant figures for on the graph sketch the variation with t

of the horizontal velocity label this sketch VH the horizontal velocity remains

constant throughout the motion so we draw the graph like this I hope you found this video helpful if you did I

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