7.5 - motion graphs

ball being thrown 2m here

what is the distance and displacement of the ball?

distance = 4m

displacement = 2m

what is displacement?

distance in a given direction

what is distance?

how much ground an object has covered during its motion

what is the gradient on a displacement-time graph?

speed at any given moment

how do you find the speed at any given moment on a displacement-time graph?

gradient = speed at any given moment

why is the gradient the speed at any given moment on a displacement-time graph?

• grad = y / x

• = s / t

• = v

how do you find the total distance travelled on a displacement-time graph?

area under the line

what is the area under the graph on a displacement-time graph?

distance

on a displacement-time graph of a ball being thrown vertically upwards, where is the maximum height the ball reaches?

here

at the peak

here

what is the displacement-time graph of a ball thrown vertically upwards?

here

• curved

• positive gradient until peak reached at maximum height

• negative gradient until displacement = 0 when ball returns to starting point

• area under graph = total distance covered

what is the distance-time graph of a ball thrown vertically upwards?

here

• curved

• positive gradient throughout

• stagnates at the maximum height

what are the similarities and differences between a displacement-time and a distance-time graph for a ball thrown vertically upwards?

here - velocity = 0 at max height, distance graph doesn’t return back to x-axis 3

for a ball thrown vertically upwards, what is the velocity at its maximum height?

zero

for a ball thrown vertically upwards, why is the maximum height the peak (on a displacement-time graph)?

because at the peak, velocity is zero (as objects stop momentarily at the stationary point) and the ball begins to fall back down, making velocity negative as displacement is negative when the ball falls

for a ball thrown vertically upwards, why is there zero gradient at the maximum height (on a distance-time graph)?

a ball momentarily stops at the turning point, so the speed is zero at maximum height. the gradient of a distance-time graph is speed, so therefore gradient is zero at maximum height

for a ball thrown vertically upwards, when is the gradient zero?

at maximum height

for a ball thrown vertically upwards, why is the gradient zero at its maximum height (on a displacement-time graph)?

a ball momentarily stops at the turning point, so the velocity is zero at maximum height. the gradient of a displacement-time graph is velocity, so therefore gradient is zero at maximum height

for a ball thrown vertically upwards, when does the graph stagnate (on a distance-time graph)?

at maximum height

for a ball thrown vertically upwards, why does the graph stagnate at the maximum height (on a distance-time graph)?

a ball momentarily stops at the turning point, so the velocity is zero at maximum height. the gradient of a displacement-time graph is velocity, so therefore the graph stagnates (has zero gradient) at maximum height

on a distance-time graph, where does the ball reach maximum height?

• when the graph stagnates

• when gradient = 0

how does a ball’s velocity vary when it is thrown vertically upwards?

• largest when it is first thrown

• lessens as it reaches maximum height

• stops completely at maximum height

• begins moving down with small velocity, accelerating from rest

• largest (equal to the initial velocity) when hitting the ground

what is a ball’s velocity the moment it is thrown upwards?

the highest it will be

what happens the ball’s velocity as it reaches its maximum height?

slows down to zero at maximum height

for a ball thrown vertically upwards, when is the velocity the greatest?

when the ball is first thrown and just before the ball hits the ground

on a displacement-time graph, is the velocity of a ball positive or negative when thrown upwards?

positive

why is the velocity of a ball positive on a displacement-time graph when thrown upwards?

because the upward direction is resolved positively, meaning displacement is positive and therefore the graph is positive

why is the velocity of a ball negative on a displacement-time graph when falling downwards?

because the downward direction is resolved negatively, meaning displacement is negative and therefore the graph is negative

on a displacement-time graph, is the velocity of a ball positive or negative when falling downwards?

negative

for a ball thrown vertically upwards, when is its velocity the smallest?

just before and after reaching maximum height

for a ball thrown vertically upwards, when is the gradient the steepest (on a displacement-time graph)?

when the ball is first thrown and just before the ball hits the ground

for a ball thrown vertically upwards, when is the gradient the shallowest (on a displacement-time graph)?

just before and after reaching maximum height

for a ball thrown vertically upwards, why is the gradient steepest when the ball is first thrown and just before the ball hits the ground (on a displacement-time graph)?

• velocity is highest when first thrown because the force throwing the ball upwards is applied

• velocity is highest just before the ball hits the ground because it has been accelerating from rest after reaching maximum height

• gradient of distance-time graph = velocity, therefore gradient is steepest when velocity is greatest

for a ball thrown vertically upwards, why is the gradient shallowest right before and after the ball reaches maximum height (on a displacement-time graph)?

• the ball slows down to complete stop at maximum height since velocity is zero at the turning point, therefore velocity is smallest just before reaching zero

• the ball, after reaching maximum height, accelerates from rest from the gravitational field strength of the earth, therefore the velocity is smallest just as it starts accelerating from rest

• gradient of displacement-time graph = velocity, therefore gradient is shallowest when velocity is smallest

for a ball thrown vertically upwards, when is the gradient the steepest (on a distance-time graph)?

when the ball is first thrown and just before the ball hits the ground

for a ball thrown vertically upwards, when is the gradient the shallowest (on a distance-time graph)?

just before and after reaching maximum height

for a ball thrown vertically upwards, why is the gradient steepest when the ball is first thrown and just before the ball hits the ground (on a distance-time graph)?

• speed is highest when first thrown because the force throwing the ball upwards is applied

• speed is highest just before the ball hits the ground because it has been accelerating from rest after reaching maximum height

• gradient of distance-time graph = speed, therefore gradient is steepest when speed is greatest

for a ball thrown vertically upwards, why is the gradient shallowest right before and after the ball reaches maximum height (on a distance-time graph)?

• the ball slows down to complete stop at maximum height since speed is zero at the turning point, therefore speed is smallest just before reaching zero

• the ball, after reaching maximum height, accelerates from rest from the gravitational field strength of the earth, therefore the speed is smallest just as it starts accelerating from rest

• gradient of distance-time graph = speed, therefore gradient is shallowest when speed is smallest

on a displacement-time graph, when is the gradient positive?

when the object is moving in a direction that is resolved positively, therefore displacement is positive, and velocity (gradient) is positive

on a displacement-time graph, when is the gradient negative?

when the object is moving in a direction that is resolved negatively, therefore displacement is negative, and velocity (gradient) is negative

on a distance-time graph, when is the gradient positive?

ALWAYS ! apart from when it is zero at the stationary points

on a distance-time graph, when is the gradient negative?

NEVER !!!! BITCH !!!!!!!

why is the gradient always positive (ignoring zero gradient at turning points) on a distance-time graph?

the gradient of a distance-time graph is speed, and as speed is a scalar quantity, it cannot have a negative value

why can the gradient vary on a displacement-time graph?

the gradient of a displacement-time graph is velocity, and as velocity is a vector quantity, it can have both positive and negative values depending on direction of motion (displacement)

why is speed always positive (ignoring zero gradient at turning points) on a distance-time graph?

speed is a scalar quantity, therefore it cannot have a negative value

why can velocity vary on a distance-time graph?

velocity is a vector quantity, therefore it can have both positive and negative values depending on direction of motion (displacement)

how can we refer to the points on a distance / distance - time graph where the gradient is zero?

turning points

why can we refer to the points on a distance / distance - time graph where the gradient is zero as the turning point?

because at this point, the ball has reached maximum height and stops moving temporarily as it changes directly to fall back down

for a ball thrown vertically upwards, what happens after reaching maximum height?

• the ball will fall back down

• the ball will accelerate from rest

• displacement is negative from this point onwards

displacement-time graph for object thrown vertically upwards here

displacement-time graph for object thrown vertically upwards

distance-time graph for object thrown vertically upwards here

distance-time graph for object thrown vertically upwards

what is velocity?

speed in a given direction

what is speed?

the rate at which an object covers distance, distance per unit time

what is the velocity-time graph for a ball thrown vertically upwards?

here

• y = velocity

• x = time

• grad = acceleration = -9.81

• area under graph = displacement

• acceleration = g, as the ball is in free fall

• acceleration = -ve, as an object thrown vertically upwards is acting against gravity

velocity-time graph for a ball thrown vertically upwards here

velocity-time graph for a ball thrown vertically upwards

what is the speed-time graph for a ball thrown vertically upwards?

here

• y = speed

• x = time

• grad = acceleration

• area under graph = displacement

• acceleration = g, as the ball is in free fall

• acceleration = -ve, as an object thrown vertically upwards is acting against gravity

• will never give negative y-values as you can’t have negative speed

speed-time graph for a ball thrown vertically upwards here

speed-time graph for a ball thrown vertically upwards

what is the gradient of a velocity-time graph?

acceleration

what is the gradient of a speed-time graph?

acceleration

what is acceleration on a velocity-time graph?

gradient

what is acceleration on a speed-time graph?

gradient

what is the y-axis on a velocity-time graph?

velocity

what is the y-axis on a speed-time graph?

speed

what is velocity on a velocity-time graph?

y-axis

what is speed on a speed-time graph?

y-axis

what is the x-axis on a velocity-time graph?

time

what is the x-axis on a speed-time graph?

time

what is time on a velocity-time graph?

x-axis

what is time on a speed-time graph?

x-axis

what is the area under a velocity-time graph?

displacement

what is the area under a speed-time graph?

displacement

what is displacement on a velocity-time graph?

area under the graph

what is displacement on a speed-time graph?

area under the graph

for a ball thrown vertically upwards, compare the velocity-time and speed-time graphs

here

similarities

• x-axis = time

• gradient = acceleration

• area under graph = displacement

differences

• a speed-time graph will never give a negative value for speed (y)

when is the gradient of a velocity-time graph positive?

when acceleration is positive

when is the gradient of a velocity-time graph negative?

when acceleration is negative

where is the maximum height reached by a ball thrown upwards on a velocity-time graph?

at the x-intercept

what is the x-intercept on a velocity-time graph?

time of zero velocity

what time does a ball thrown vertically upwards reach maximum height?

t = u / g

t = u / g

time when a ball thrown vertically upwards reaches maximum height

why is the time a ball thrown vertically upwards reaches maximum height t = u / g?

idk

where is the maximum height reached by a ball thrown upwards on a speed-time graph?

when the graph touches the x-axis

why is the maximum height reached by a ball thrown upwards on a speed-time graph the point where the graph touches the x-axis?

the ball stops moving momentarily when reaching maximum height, therefore speed is zero at the maximum height. as y = speed on a velocity-time graph, the graph touches the x-axis when speed = zero, therefore the ball reaches maximum height when it touches the x-axis

what is the difference between a velocity-time graph?

a speed-time graph will never give negative values for speed

here comparison velocity time and speed time for ball thrown upwards

why will a speed-time graph never give negative values for speed (y)?

speed is a scalar quantity, meaning it only has a magnitude and not a direction. it doesn’t matter which direction the object is travelling as it does not account for displacement, therefore it will never give a negative value for speed as the object cannot have negative speed

for an object in free fall, what is the gradient of a velocity-time equal to?

negative gravitational field strength, -9.81 ms^-2

for an object in free fall, why is the gradient negative gravitational field strength, -9.81 ms^-2 ?

an object in free fall is only acted on by the gravitational field strength, g. an object thrown upwards is acting against gravity, therefore it’s acceleration is -g

for an object thrown vertically upwards, is the gradient of a velocity-time graph positive or negative?

negative

for an object thrown vertically upwards, is the acceleration positive or negative?

negative

for an object thrown vertically upwards, is the gradient of a velocity-time graph constant?

yes

for an object thrown vertically upwards, why is the gradient of a velocity-time graph constant?

an object thrown vertically upwards is moving in free fall, therefore it is only acted on by the gravitational field strength, g, which is constant. the gradient of a velocity-time graph is acceleration, so the gradient is constant

for an object thrown vertically upwards, is the acceleration constant?

yes

for an object thrown vertically upwards, why is the acceleration constant?

an object thrown vertically upwards is moving in free fall, therefore it is only acted on by the gravitational field strength, g, which is constant

for an object thrown vertically upwards, why is gradient of a velocity-time graph negative?

an object thrown vertically upwards is moving in free fall, therefore it is only acted on by the gravitational field strength, g. an object thrown upwards is acting against gravity, therefore it’s acceleration is -g. the gradient of a velocity-time graph is acceleration, so the gradient is negative

for an object thrown vertically upwards, why is the acceleration negative?

an object thrown vertically upwards is moving in free fall, therefore it is only acted on by the gravitational field strength, g. an object thrown upwards is acting against gravity, therefore it’s acceleration is -g

for a velocity-time graph depicting a ball being thrown vertically upwards, is the displacement positive or negative before the maximum height is reached?

positive

for a velocity-time graph depicting a ball being thrown vertically upwards, is the displacement positive or negative after the maximum height is reached?

negative

for a velocity-time graph depicting a ball being thrown vertically upwards, is the area positive or negative before the maximum height is reached?

positive

for a velocity-time graph depicting a ball being thrown vertically upwards, is the area positive or negative after the maximum height is reached?

negative

for a velocity-time graph depicting a ball being thrown vertically upwards, why is the area positive before the maximum height is reached?

before the ball has reached maximum height, it it moving upwards. up is resolved positively, so displacement is positively when the ball is thrown up. displacement is area under the graph, therefore area is positive as displacement is positive before maximum height is reached