PowerPoint Presentation
Kinematics of a particle moving in a straight line
IntroductionThis chapter you will learn the SUVAT equations
These are the foundations of many of the Mechanics topics
You will see how to use them to use many types of problem
involving motionTeaChings for Exercise 2AKinematics of a Particle
moving in a Straight LineYou will begin by learning two of the
SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AReplace with the
appropriate letters. Change in velocity = final velocity initial
velocityMultiply by tAdd uThis is the usual form!Replace with the
appropriate letters
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AYou need to consider using
negative numbers in some casesPQPositive
directionO4m3m2.5ms-16ms-1If we are measuring displacements from O,
and left to right is the positive directionFor particle P:For
particle Q:The particle is to the left of the point O, which is the
negative directionThe particle is moving at 2.5ms-1 in the positive
directionThe particle is to the right of the point O, which is the
positive directionThe particle is moving at 6ms-1 in the negative
direction
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AA particle is moving in a
straight line from A to B with constant acceleration 3ms-2. Its
speed at A is 2ms-1 and it takes 8 seconds to move from A to B.
Find:The speed of the particle at BThe distance from A to
BAB2ms-1Start with a diagramWrite out suvat and fill in what you
knowFor part a) we need to calculate v, and we know u, a and tFill
in the values you knowRemember to include units!You always need to
set up the question in this way. It makes it much easier to figure
out what equation you need to use (there will be more to learn than
just these two!)
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AA particle is moving in a
straight line from A to B with constant acceleration 3ms-2. Its
speed at A is 2ms-1 and it takes 8 seconds to move from A to B.
Find:The speed of the particle at B 26ms-1The distance from A to
BAB2ms-1For part b) we need to calculate s, and we know u, v and
tFill in the values you knowShow calculationsRemember the
units!
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AA cyclist is travelling
along a straight road. She accelerates at a constant rate from a
speed of 4ms-1 to a speed of 7.5ms-1 in 40 seconds. Find:The
distance travelled over this 40 secondsThe acceleration over the 40
seconds4ms-17.5ms-1Draw a diagram (model the cyclist as a
particle)Write out suvat and fill in what you knowWe are
calculating s, and we already know u, v and tSub in the values you
knowRemember units!
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AA cyclist is travelling
along a straight road. She accelerates at a constant rate from a
speed of 4ms-1 to a speed of 7.5ms-1 in 40 seconds. Find:The
distance travelled over this 40 seconds 230mThe acceleration over
the 40 seconds4ms-17.5ms-1Draw a diagram (model the cyclist as a
particle)Write out suvat and fill in what you knowFor part b, we
are calculating a, and we already know u, v and tSub in the values
you knowSubtract 4Divide by 40
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AA particle moves in a
straight line from a point A to B with constant deceleration of
1.5ms-2. The speed of the particle at A is 8ms-1 and the speed of
the particle at B is 2ms-1. Find:The time taken for the particle to
get from A to BThe distance from A to B8ms-12ms-1Draw a
diagramWrite out suvat and fill in what you knowAs the particle is
decelerating, a is negativeSub in the values you knowSubtract
8Divide by -1.5AB
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AA particle moves in a
straight line from a point A to B with constant deceleration of
1.5ms-2. The speed of the particle at A is 8ms-1 and the speed of
the particle at B is 2ms-1. Find:The time taken for the particle to
get from A to B 4 secondsThe distance from A to B8ms-12ms-1Draw a
diagramWrite out suvat and fill in what you knowAs the particle is
decelerating, a is negativeSub in the values you knowCalculate the
answer!AB
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AAfter reaching B the
particle continues to move along the straight line with the same
deceleration. The particle is at point C, 6 seconds after passing
through A. Find:The velocity of the particle at CThe distance from
A to C8ms-12ms-1ABC?Update the diagramWrite out suvat using points
A and CSub in the valuesWork it out!As the velocity is negative,
this means the particle has now changed direction and is heading
back towards A! (velocity has a direction as well as a
magnitude!)The velocity is 1ms-1 in the direction C to A
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AAfter reaching B the
particle continues to move along the straight line with the same
deceleration. The particle is at point C, 6 seconds after passing
through A. Find:The velocity of the particle at C - -1ms-1The
distance from A to C8ms-12ms-1ABC?Update the diagramWrite out suvat
using points A and CSub in the valuesWork it out!It is important to
note that 21m is the distance from A to C only The particle was
further away before it changed direction, and has in total
travelled further than 21m
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AA car moves from traffic
lights along a straight road with constant acceleration. The car
starts from rest at the traffic lights and 30 seconds later passes
a speed trap where it is travelling at 45 kmh-1. Find:The
acceleration of the carThe distance between the traffic lights and
the speed-trap.0ms-145kmh-1LightsTrapStandard units to use are
metres and seconds, or kilometres and hoursIn this case, the time
is in seconds and the speed is in kilometres per hourWe need to
change the speed into metres per second first!Draw a
diagramMultiply by 1000 (km to m)Divide by 3600 (hours to
seconds)
Kinematics of a Particle moving in a Straight LineYou will begin
by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AA car moves from traffic
lights along a straight road with constant acceleration. The car
starts from rest at the traffic lights and 30 seconds later passes
a speed trap where it is travelling at 45 kmh-1. Find:The
acceleration of the carThe distance between the traffic lights and
the speed-trap.0ms-145kmh-1LightsTrapDraw a diagram= 12.5ms-1Write
out suvat and fill in what you knowSub in the valuesDivide by 30You
can use exact answers!
15Kinematics of a Particle moving in a Straight LineYou will
begin by learning two of the SUVAT equations
s = Displacement (distance)u = Starting (initial) velocityv =
Final velocitya = Accelerationt = Time2AA car moves from traffic
lights along a straight road with constant acceleration. The car
starts from rest at the traffic lights and 30 seconds later passes
a speed trap where it is travelling at 45 kmh-1. Find:The
acceleration of the carThe distance between the traffic lights and
the speed-trap.0ms-145kmh-1LightsTrapDraw a diagram= 12.5ms-1Write
out suvat and fill in what you knowSub in valuesWork it out!
16TeaChings for Exercise 2BKinematics of a Particle moving in a
Straight LineYou can also use 3 more formulae linking different
combination of SUVAT, for a particle moving in a straight line with
constant acceleration2BSubtract uDivide by aReplace t with the
expression aboveMultiply numerators and denominatorsMultiply by
2aAdd u2This is the way it is usually written!
Kinematics of a Particle moving in a Straight LineYou can also
use 3 more formulae linking different combination of SUVAT, for a
particle moving in a straight line with constant
acceleration2BReplace v with u + atGroup terms on the
numeratorDivide the numerator by 2Multiply out the bracket
Kinematics of a Particle moving in a Straight LineYou can also
use 3 more formulae linking different combination of SUVAT, for a
particle moving in a straight line with constant
acceleration2BSubtract atReplace u with v - at from aboveMultiply
out the bracketGroup up the at2 terms
Kinematics of a Particle moving in a Straight LineYou can also
use 3 more formulae linking different combination of SUVAT, for a
particle moving in a straight line with constant acceleration2BA
particle is moving in a straight line from A to B with constant
acceleration 5ms-2. The velocity of the particle at A is 3ms-1 in
the direction AB. The velocity at B is 18ms-1 in the same
direction. Find the distance from A to B.3ms-118ms-1ABDraw a
diagramWrite out suvat with the information givenReplace v, u and
aWork out termsSubtract 9Divide by 10We are calculating s, using v,
u and a
Kinematics of a Particle moving in a Straight LineYou can also
use 3 more formulae linking different combination of SUVAT, for a
particle moving in a straight line with constant acceleration2BA
car is travelling along a straight horizontal road with a constant
acceleration of 0.75ms-2. The car is travelling at 8ms-1 as it
passes a pillar box. 12 seconds later the car passes a lamp post.
Find:The distance between the pillar box and the lamp postThe speed
with which the car passes the lamp post8ms-1Pillar BoxLamp PostDraw
a diagramWrite out suvat with the information givenWe are
calculating s, using u, a and tReplace u, a and tCalculate
Kinematics of a Particle moving in a Straight LineYou can also
use 3 more formulae linking different combination of SUVAT, for a
particle moving in a straight line with constant acceleration2BA
car is travelling along a straight horizontal road with a constant
acceleration of 0.75ms-2. The car is travelling at 8ms-1 as it
passes a pillar box. 12 seconds later the car passes a lamp post.
Find:The distance between the pillar box and the lamp post 150mThe
speed with which the car passes the lamp post8ms-1Pillar BoxLamp
PostDraw a diagramWrite out suvat with the information givenWe are
calculating v, using u, a and tReplace u, a and tCalculateOften you
can use an answer you have calculated later on in the same
question. However, you must take care to use exact values and not
rounded answers!
Kinematics of a Particle moving in a Straight LineYou can also
use 3 more formulae linking different combination of SUVAT, for a
particle moving in a straight line with constant acceleration2BA
particle is moving in a straight horizontal line with constant
deceleration 4ms-2. At time t = 0 the particle passes through a
point O with speed 13ms-1, travelling to a point A where OA = 20m.
Find:The times when the particle passes through AThe total time the
particle is beyond AThe time taken for the particle to return to
O13ms-1OADraw a diagramWrite out suvat with the information givenWe
are calculating t, using s, u and aReplace s, u and aSimplify
termsRearrange and set equal to 0Factorise (or use the quadratic
formula)We have 2 answers. As the acceleration is negative, the
particle passes through A, then changes direction and passes
through it again!
Kinematics of a Particle moving in a Straight LineYou can also
use 3 more formulae linking different combination of SUVAT, for a
particle moving in a straight line with constant acceleration2BA
particle is moving in a straight horizontal line with constant
deceleration 4ms-2. At time t = 0 the particle passes through a
point O with speed 13ms-1, travelling to a point A where OA = 20m.
Find:The times when the particle passes through A 2.5 and 4
secondsThe total time the particle is beyond AThe time taken for
the particle to return to O13ms-1OADraw a diagramWrite out suvat
with the information givenWe are calculating t, using s, u and aThe
particle passes through A at 2.5 seconds and 4 seconds, so it was
beyond A for 1.5 seconds
Kinematics of a Particle moving in a Straight LineYou can also
use 3 more formulae linking different combination of SUVAT, for a
particle moving in a straight line with constant acceleration2BA
particle is moving in a straight horizontal line with constant
deceleration 4ms-2. At time t = 0 the particle passes through a
point O with speed 13ms-1, travelling to a point A where OA = 20m.
Find:The times when the particle passes through A 2.5 and 4
secondsThe total time the particle is beyond A 1.5 secondsThe time
taken for the particle to return to O13ms-1OADraw a diagramWrite
out suvat with the information givenThe particle returns to O when
s = 0Replace s, u and aSimplifyRearrangeFactoriseThe particle is at
O when t = 0 seconds (to begin with) and is at O again when t = 6.5
seconds
Kinematics of a Particle moving in a Straight LineYou can also
use 3 more formulae linking different combination of SUVAT, for a
particle moving in a straight line with constant acceleration2BA
particle is travelling along the x-axis with constant deceleration
2.5ms-2. At time t = O, the particle passes through the origin,
moving in the positive direction with speed 15ms-1. Calculate the
distance travelled by the particle by the time it returns to the
origin.15ms-1OXDraw a diagramThe total distance travelled will be
double the distance the particle reaches from O (point X)
At X, the velocity is 0Replace v, u and aSimplifyAdd 5sDivide by
545m is the distance from O to X. Double it for the total distance
travelledWe are calculating s, using u, v and a
TeaChings for Exercise 2CKinematics of a Particle moving in a
Straight LineYou can use the formulae for constant acceleration to
model an object moving vertically in a straight line under the
influence of gravity
Gravity causes objects to fall to the earth! (as you probably
already know!)
The acceleration caused by gravity is constant (if you ignore
air resistance)
This means the acceleration will be the same, regardless of the
size of the object
On Earth, the acceleration due to gravity is 9.8ms-2, correct to
2 significant figures.
When solving problems involving vertical motion you must
carefully consider the direction. As gravity acts in a downwards
direction:An object thrown downwards will have an acceleration of
9.8ms-2An object thrown upwards will have an acceleration of
-9.8ms-2
The time of flight is the length of time an object spends in the
air. The speed of projection is another name for the objects
initial speed (u)2C
Kinematics of a Particle moving in a Straight LineYou can use
the formulae for constant acceleration to model an object moving
vertically in a straight line under the influence of gravity2CA
ball is projected vertically upwards from a point O with a speed of
12ms-1. Find:The greatest height reached by the ballThe total time
the ball is in the air12ms-10ms-1Draw a diagramAt its highest
point, the velocity of the ball is 0ms-1As the ball has been
projected upwards, gravity is acting in the opposite direction and
hence the acceleration is negativeReplace v, u and aSimplifyAdd
19.6sDivide and round to 2sf (since gravity has been given to
2sf)We are calculating s, using u, v and a
Kinematics of a Particle moving in a Straight LineYou can use
the formulae for constant acceleration to model an object moving
vertically in a straight line under the influence of gravity2CA
ball is projected vertically upwards from a point O with a speed of
12ms-1. Find:The greatest height reached by the ball 7.4mThe total
time the ball is in the air12ms-10ms-1Draw a diagramFor the total
time the ball is in the air, the displacement (s) will be 0Also, we
will not know v (yet!) when the ball strikes the groundWe are
calculating t, using s, u and aReplace s, u and aFactoriseChoose
the appropriate answer!So the ball will be in the air for 2.4
seconds
Kinematics of a Particle moving in a Straight LineYou can use
the formulae for constant acceleration to model an object moving
vertically in a straight line under the influence of gravity2CA
book falls off the top shelf of a bookcase. The shelf is 1.4m above
the ground. Find:The time it takes the book to reach the floorThe
speed with which the book strikes the floor0ms-1Draw a
diagram1.4mThe books initial speed will be 0 as it has not been
projected to begin withAs the books initial movement is downwards,
we take the acceleration due to gravity as positiveWe are
calculating t, using s, u and aReplace s, u and aSimplifyDivide by
4.9Find the positive square root
Kinematics of a Particle moving in a Straight LineYou can use
the formulae for constant acceleration to model an object moving
vertically in a straight line under the influence of gravity2CA
book falls off the top shelf of a bookcase. The shelf is 1.4m above
the ground. Find:The time it takes the book to reach the floor 0.53
secondsThe speed with which the book strikes the floor0ms-1Draw a
diagram1.4mThe books initial speed will be 0 as it has not been
projected to begin withAs the books initial movement is downwards,
we take the acceleration due to gravity as positiveWe are
calculating v, using s, u and aReplace s, u and aCalculateFind the
positive square root
Kinematics of a Particle moving in a Straight LineYou can use
the formulae for constant acceleration to model an object moving
vertically in a straight line under the influence of gravity2CA
ball is projected upwards from a point X which is 7m above the
ground, with initial speed 21ms-1. Find the time of flight of the
ball.21ms-17mDraw a diagramThe balls flight will last until it hits
the groundWe want the ball to be 7m lower than it starts (in the
negative direction)Hence, s = -7The ball is projected upwards, so
the acceleration due to gravity is negativeWe are calculating t,
using s, u and aReplace s, u and aSimplifyRearrange and set equal
to 0We will need the quadratic formula here, so write down a, b and
c
Kinematics of a Particle moving in a Straight LineYou can use
the formulae for constant acceleration to model an object moving
vertically in a straight line under the influence of gravity2CA
ball is projected upwards from a point X which is 7m above the
ground, with initial speed 21ms-1. Find the time of flight of the
ball.21ms-17mDraw a diagramThe balls flight will last until it hits
the groundWe want the ball to be 7m lower than it starts (in the
negative direction)Hence, s = -7The ball is projected upwards, so
the acceleration due to gravity is negativeReplace a, b and c
(using brackets!)Calculate and be careful with any negatives in the
previous step!)
Kinematics of a Particle moving in a Straight LineYou can use
the formulae for constant acceleration to model an object moving
vertically in a straight line under the influence of gravity2CA
particle is projected vertically upwards from a point O with
initial speed u ms-1. The greatest height reached by the particle
is 62.5m above the ground. Find:The speed of projectionThe total
time for which the ball is 50m or more above the groundu
ms-162.5mDraw a diagramThe maximum height is 62.5m At this point
the balls velocity is 0ms-1The ball is projected upwards, so the
acceleration due to gravity is negativeWe are calculating u, using
s, v and aReplace v, a and sSimplifyRewriteFind the positive square
root
Kinematics of a Particle moving in a Straight LineYou can use
the formulae for constant acceleration to model an object moving
vertically in a straight line under the influence of gravity2CA
particle is projected vertically upwards from a point O with
initial speed u ms-1. The greatest height reached by the particle
is 62.5m above the ground. Find:The speed of projection 35ms-1The
total time for which the ball is 50m or more above the groundu
ms-162.5mDraw a diagramThe ball will pass the 50m mark twice we
need to find these two times!50mWe are calculating t, using s, u
and aReplace s, u and aSimplifyRearrange, and set equal to 0We will
need the quadratic formula, and hence a, b and c
37Kinematics of a Particle moving in a Straight LineYou can use
the formulae for constant acceleration to model an object moving
vertically in a straight line under the influence of gravity2CA
particle is projected vertically upwards from a point O with
initial speed u ms-1. The greatest height reached by the particle
is 62.5m above the ground. Find:The speed of projection 35ms-1The
total time for which the ball is 50m or more above the groundu
ms-162.5mDraw a diagramThe ball will pass the 50m mark twice we
need to find these two times!50mWe are calculating t, using s, u
and aSub these into the Quadratic formulaWe get the two times the
ball passes the 50m markCalculate the difference between these
times!
38Kinematics of a Particle moving in a Straight LineYou can use
the formulae for constant acceleration to model an object moving
vertically in a straight line under the influence of gravity2CA
ball, A, falls vertically from rest from the top of a tower 63m
high. At the same time as A begins to fall, another ball, B, is
projected vertically upwards from the bottom of the tower with
velocity 21ms-1. The balls collide. Find the height at which this
happens.
63ms1s221ms-1Draw a diagramIn this case we need to consider each
ball separately.We can call the two distances s1 and s2The time
will be the same for both when they collide, so we can just use
tMake sure that acceleration is positive for A as it is travelling
downwards and negative for B as it is travelling upwardsSub in s,
u, a and t for Ball BSimplifySub in s, u, a and t for Ball
ASimplify39Kinematics of a Particle moving in a Straight LineYou
can use the formulae for constant acceleration to model an object
moving vertically in a straight line under the influence of
gravity2CA ball, A, falls vertically from rest from the top of a
tower 63m high. At the same time as A begins to fall, another ball,
B, is projected vertically upwards from the bottom of the tower
with velocity 21ms-1. The balls collide. Find the height at which
this happens.
63ms1s221ms-1Draw a diagramIn this case we need to consider each
ball separately.We can call the two distances s1 and s2The time
will be the same for both when they collide, so we can just use
tMake sure that acceleration is positive for A as it is travelling
downwards and negative for B as it is travelling upwards1)2)Add the
two equations together (this cancels the 4.9t2 terms)s1 + s2 must
be the height of the tower (63m)Divide by 21So the balls collide
after 3 seconds40Kinematics of a Particle moving in a Straight
LineYou can use the formulae for constant acceleration to model an
object moving vertically in a straight line under the influence of
gravity2CA ball, A, falls vertically from rest from the top of a
tower 63m high. At the same time as A begins to fall, another ball,
B, is projected vertically upwards from the bottom of the tower
with velocity 21ms-1. The balls collide. Find the height at which
this happens.
63ms1s221ms-1Draw a diagramIn this case we need to consider each
ball separately.We can call the two distances s1 and s2The time
will be the same for both when they collide, so we can just use
tMake sure that acceleration is positive for A as it is travelling
downwards and negative for B as it is travelling upwards2)Sub in t
= 3 (we use this equation since s2 is the height above the
ground)41TeaChings for Exercise 2DKinematics of a Particle moving
in a Straight LineYou can represent the motion of an object on a
speed-time graph, distance-time graph or an acceleration-time
graph2D
OuvtInitial velocityFinal velocityTime takenv - utOn a
speed-time graph, the gradient of a section is its
acceleration!vutOn a speed-time graph, the Area beneath it is the
distance covered!Kinematics of a Particle moving in a Straight
LineYou can represent the motion of an object on a speed-time
graph, distance-time graph or an acceleration-time graph
Gradient of a speed-time graph = Acceleration over that
period
Area under a speed-time graph = distance travelled during that
period2D
A car accelerates uniformly at 5ms-2 from rest for 20 seconds.
It then travels at a constant speed for the next 40 seconds, then
decelerates uniformly for the final 20 seconds until it is at rest
again.Draw an acceleration-time graph for this informationDraw a
distance-time graph for this information204060805Acceleration
(ms-2)0-5For now, we assume the rate of acceleration jumps between
different ratesTime (s)20406080Time (s)As the speed increases the
curve gets steeper, but with a constant speed the curve is
straight. Finally the curve gets less steep as deceleration takes
placeDistance (m)Kinematics of a Particle moving in a Straight
LineYou can represent the motion of an object on a speed-time
graph, distance-time graph or an acceleration-time graph
Gradient of a speed-time graph = Acceleration over that
period
Area under a speed-time graph = distance travelled during that
period2D
The diagram below shows a speed-time graph for the motion of a
cyclist moving along a straight road for 12 seconds. For the first
8 seconds, she moves at a constant speed of 6ms-1. She then
decelerates at a constant rate, stopping after a further 4 seconds.
Find:The distance travelled by the cyclistThe rate of deceleration
of the cyclistv(ms-1)t(s)068128126Sub in the appropriate values for
the trapezium aboveCalculateKinematics of a Particle moving in a
Straight LineYou can represent the motion of an object on a
speed-time graph, distance-time graph or an acceleration-time
graph
Gradient of a speed-time graph = Acceleration over that
period
Area under a speed-time graph = distance travelled during that
period2D
The diagram below shows a speed-time graph for the motion of a
cyclist moving along a straight road for 12 seconds. For the first
8 seconds, she moves at a constant speed of 6ms-1. She then
decelerates at a constant rate, stopping after a further 4 seconds.
Find:The distance travelled by the cyclist 60mThe rate of
deceleration of the cyclistv(ms-1)t(s)068124-6Sub in the
appropriate values for the trapezium aboveCalculateKinematics of a
Particle moving in a Straight LineYou can represent the motion of
an object on a speed-time graph, distance-time graph or an
acceleration-time graph
Gradient of a speed-time graph = Acceleration over that
period
Area under a speed-time graph = distance travelled during that
period2D
A particle moves along a straight line. It accelerates uniformly
from rest to a speed of 8ms-1 in T seconds. The particle then
travels at a constant speed for 5T seconds. It then decelerates to
rest uniformly over the next 40 seconds.Sketch a speed-time graph
for this motionGiven that the particle travels 600m, find the value
of TSketch an acceleration-time graph for this
motionv(ms-1)t(s)08T5T405T86T + 40Sub in valuesSimplify
fractionDivide by 8Subtract 20Divide by 5.5Kinematics of a Particle
moving in a Straight LineYou can represent the motion of an object
on a speed-time graph, distance-time graph or an acceleration-time
graph
Gradient of a speed-time graph = Acceleration over that
period
Area under a speed-time graph = distance travelled during that
period2D
A particle moves along a straight line. It accelerates uniformly
from rest to a speed of 8ms-1 in T seconds. The particle then
travels at a constant speed for 5T seconds. It then decelerates to
rest uniformly over the next 40 seconds.Sketch a speed-time graph
for this motionGiven that the particle travels 600m, find the value
of T 10 secondsSketch an acceleration-time graph for this
motionv(ms-1)t(s)08T5T405010First sectionLast
sectiont(s)a(ms-2)204060801000.8-0.2Kinematics of a Particle moving
in a Straight LineYou can represent the motion of an object on a
speed-time graph, distance-time graph or an acceleration-time
graph
Gradient of a speed-time graph = Acceleration over that
period
Area under a speed-time graph = distance travelled during that
period2D
A car C is moving along a straight road with constant speed
17.5ms-1. At time t = 0, C passes a lay-by. Also at time t = 0, a
second car, D, leaves the lay-by. Car D accelerates from rest to a
speed of 20ms-1 in 15 seconds and then maintains this speed. Car D
passes car C at a road sign.Sketch a speed-time graph to show the
motion of both carsCalculate the distance between the lay-by and
the road signv(ms-1)t(s)02017.515CDAt the road sign, the cars have
covered the same distance in the same time
We need to set up simultaneous equations using s and t
Let us call the time when the areas are equal TT17.5TT - 1520Sub
in valuesSub in valuesSimplify fractionMultiply bracketKinematics
of a Particle moving in a Straight LineYou can represent the motion
of an object on a speed-time graph, distance-time graph or an
acceleration-time graph
Gradient of a speed-time graph = Acceleration over that
period
Area under a speed-time graph = distance travelled during that
period2D
A car C is moving along a straight road with constant speed
17.5ms-1. At time t = 0, C passes a lay-by. Also at time t = 0, a
second car, D, leaves the lay-by. Car D accelerates from rest to a
speed of 20ms-1 in 15 seconds and then maintains this speed. Car D
passes car C at a road sign.Sketch a speed-time graph to show the
motion of both carsCalculate the distance between the lay-by and
the road signv(ms-1)t(s)02017.515CDAt the road sign, the cars have
covered the same distance in the same time
We need to set up simultaneous equations using s and t
Let us call the time when the areas are equal TTSubtract
17.5TAdd 150Divide by 2.5Sub in TCalculate!Set these equations
equal to each other!SummaryThis chapter we have seen how to solve
problems involving the motion of a particle in a straight line,
with constant acceleration
We have extended the problems to vertical motion involving
gravity
We have also seen how to solve problems involving the motion of
two particles
We have also used graphs to solve some more complicated
problems
