distance displacement - 1634maths · pdf fileit is usually taught under the ... since this is...

DISPLACEMENT, VELOCITY AND ACCELERATION

• Mechanics is the study of the effects of forces on bodies. It is usually taught under the following two main topics:

! (i) dynamics (when the body is in the state of motion)! (ii) statics (when the body is in the state of rest)

• Kinematics is a part of dynamic where a particle moves (in a straight line) without an external forces is applied.

• A few quantities are required to describe the motion of a particle. These physical quantities are distance, displacement, speed, velocity and acceleration.

For example,

Taking point O as the reference point,

(i) the displacement of the particle at A =

(ii)the displacement of the particle at B =

(iii)the displacement of the particle at C =

(iv)the total distance travelled by the particle from O to C =

! KINEMATICS OF MOTION IN A STRAIGHT LINE

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DISTANCE DISPLACEMENT• the length of a given path moved by

a particle• S.I Unit is metre (m)• Scalar quantity because it involves

magnitude only

• the position of a particle relative to a reference point

• to specify a displacement of a particle P at any instant, we need:

1. Distance from the reference point of P2. The direction from the reference point to P

• vector quantity because it involves both magnitude and direction

path of motion

path of motion

7m

4m5m

O A BC

• Speed of a particle is the rate of change of distance with respect to time. Speed is a scalar quantity because it involves only magnitude. Metres per second (m/s or ) is the S.I unit for speed. When a particle moves with varying speed, we often speak of average speed which is defined by

Example 1A particle moves in a straight line with a uniform speed of 5 for the first 3 seconds and 6 for the next 4 seconds. Find the average speed of the particle.

• Velocity is defined as the rate of change of displacement with respect to time. A particle is said to move with uniform velocity if it moves with uniform speed and the direction of motion does not change.

• When the velocity of a particle changes, the particle is said to move with acceleration. Acceleration is the rate of velocity with respect to time. The S.I unit for acceleration is metres per second square ( or

• The acceleration is zero when the particle moves with constant velocity.


2

ms−1

AVERAGE SPEED = Total Distance TravelledTotal Time Taken

ms−1 ms−1

AVERAGE VELOCITY = Change in DisplacementTime Taken

ms−2 m/s2

AVERAGE ACCELERATION = Change in VelocityTime Taken

MOTION WITH NON-UNIFORM ACCELERATION

If the displacement, s, is a function of t time, e.g

Then velocity, rate of change of distance,

And acceleration rate of change of velocity,

Example 2A vehicle moves in a straight line so that its acceleration after t seconds is given by . Initially it has velocity of . Find an expression for the displacement s metres of the vehicle from its starting point.


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DIFFERENTIATION

S V av =

ds

dt

s =

�v dt

a =d2s

dt2=

dv

dt

v =

�a dt

INTEGRATION

Maximum Velocity / Constant Velocity: a = 0Instantaneous Rest: v = 0

Return to Original Position / Starting Position: s = 0

a = 2t

20 ms−1

s(t) = 3t2 + 8t

Example 3A particle moves in a straight line with a velocity given by , where t is the time in seconds after passing through a fixed point O on the line. Find(a) the acceleration of P after 4 seconds, (b) the displacement of P from O after 6 seconds,(c) the total distance moved by the particle in the first 6 seconds of the motion

Example 4A particle Q moves in a straight line so that its distance, s m, from a fixed point O on the line is given by where t is the time in seconds after passing O. Find(a) the velocity and speed of Q after 5 seconds(b) the distance of Q from O when it is instantaneously at rest after passing through O(c) the total distance travelled during the first 5 seconds

Example 5A particle moves along the x-axis. It passes the origin O with a velocity of and its acceleration after t seconds is given by . Calculate(a) the velocity of the particle when (b) the maximum velocity of the particle(c) the displacement of the particle from O when it is at rest(d) the average speed during the first 12 seconds


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v ms−1 v = 4− t2

s = 6t2 − t3

8 ms−1 a ms−2

a = 3− t

t = 2

DISPLACEMENT-TIME GRAPH ( s-t graph)

• The motion of a particle can be represented graphically by plotting the displacement of the particle from a given point verses the corresponding time. The graph obtained is called the displacement-time graph or the s-t graph. The graph is

(i) a straight line if the particle moves with uniform velocity(ii)a curve if the particle moves with non-uniform velocity

• The graph explains

(i) Particle started from a fixed point O at time t = 0 (initial time) and moves in the positive direction (given by the positive gradient of the curve) away from O.

(ii)At , it reaches the maximum distance and having velocity of zero which implies that the particle is momentarily at rest before turning in the opposite direction (negative direction) towards O.

(iii)Then it returns to O at (iv)It moves beyond O between and in the negative direction.

• Relating the movement of the particle (in a straight line graph) in real situation with the graph above, we have:

• For any straight line displacement-time graph,

• When the graph is NOT a straight line, the gradient of the curve at any point gives the velocity of the object at that instant.


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s(m)

t(s)0

ds

dt> 0

ds

dt< 0

ds

dt= 0

s1

−s2

t1 t2

t3

s(m)

s1t1

t2.

t2 t3

GRADIENT = Change in DisplacementTime Taken to Change

= VELOCITY ⇒ v =ds

dt

Example 1The graph shows the relation between distance and displacement s (in metres) and time t (in seconds) of a body.

(a) What happened in the different stages of the journey?(b) Draw the motion of the body in a straight line graph (in real situation)

(a) From the graph, we note the following facts:

(i) The starting or initial position A of the body is ____________ from fixed point O. The initial velocity is given by the ____________ at A and is positive. This means that the body is moving away from O in the positive direction.

(ii) From A to B, the body moves further away from O but its velocity is _______________ as the gradient is decreasing.

(iii)At B, 7m from O and 5s from the start, the gradient is ____________ (gradient at turning point) which means that the velocity is __________. This implies that the body is momentarily at ___________.

(iv) From B to C, the gradient is negative. Hence the velocity is _____________ that is the body is moving ______________ from O and arrives there after 12s from the start.

(v) The body reaches position D which is __________ from O in the opposite direction to B after 18s and is again momentarily at rest.

(vi) From D to E, the body is now moving in the positive direction and reaches O again after ____________ from the start.

(b)


6

7

3

s(m)

50

−5

12

C

A

B

D

Et(s)

18

20

Example 2A boy walks at for 3 hours. Draw his displacement-time graph.

Time (h) Dist (km)

0

1

2

3

Example 3The graph shows the displacement (in km) of a cyclist from a point A, plotted against time (in hours).

(a) What assumptions have been made in the graph drawn?(b) What happened in the different stages of the journey?

(a) Since this is a displacement-time graph, the gradient at each point represents the velocity of the cyclist at that time. The sections of the graph are straight and so we assumed that the velocity of the cyclist was uniform during each stage.

(b) Examining each section of the graph in turn,

! AB - Journey from A to B took ____________ hours and the displacement was ___________ km. ! ! Velocity = __________

! BC - Gradient of this part of the graph is ____________. The cyclist had __________ velocity and so he ! ! must have had an hourʼs rest.

! CD - Gradient of this section of the graph is ______________________ and therefore, the cyclistʼs ! ! velocity = ____________


7

4 kmh−1

s(km)

time(h)

20

10

A0

B C

D

2 3 6

VELOCITY-TIME GRAPH ( v-t graph)

• Acceleration and distance travelled by a particle can be obtained from its velocity-time graph. A particle is said to move with uniform (or constant) acceleration if the change in velocity remains the same for any equal time interval.

• For motion with uniform acceleration, the velocity-time graph is a straight line graph. Below shows 3 types of velocity-time graphs for motion with constant acceleration during the interval from

(a) Motion with zero acceleration (uniform velocity)

(b) Motion with positive uniform acceleration

(c) Motion with negative uniform acceleration (retardation)

• In general, for a particle moving with constant acceleration, the velocity-time graph is a straight-line and acceleration = gradient of the line.


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velocity

timet1 t20

velocity

timet1 t20

velocity

timet1 t20

v1 = v2

v1

v2

v2

v1

• For any straight-line velocity-time graph,

• The graph explains:

(i) At t = 0, velocity (initial) = (ii) From t = 0 to , the particle is accelerating whereas from to is decelerating (retarding).(iii) At , the particle is at rest(iv) From to , the particle is moving in the opposite direction(v) At , the particle have maximum velocity.(vi) The area under the graph gives the total distance, that is,

! ! TOTAL DISTANCE =

Example 1The diagram shows the velocity-time graph of a particle moving in a straight line.

(a) Describe very briefly the motion of the particle

(b) If the particle has returned to its starting point ! after time , what deduction can be made ! about the regions marked A, B and C?


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GRADIENT = Change in VelocityTime Taken to Change

= ACCELERATION ⇒ a =dv

dt

TOTAL DISTANCE = AREA UNDER THE GRAPH� t2

0v dt+

� t3

t2

v dt

t(s)

v(ms−1)

v1

a = 0

a > 0a < 0

t1 t2

t3

0

v1t = t1

t2

t1 t3

t3t2t1

t3

velocity

time

AB

C0 t1 t2

t3

Example 2A car accelerates at a uniform rate of for 4 seconds. It then travels at constant velocity for 5 seconds before slowing down to stop at a constant rate of . Draw a v-t graph for the motion and from it, find the total distance travelled in the journey.

Example 3A car accelerates uniformly from rest, at a rate of to speed of . It then continue moving at for T seconds before decelerating uniformly to rest in a further 10 seconds.

(a) Sketch the velocity-time graph for motion of the car.(b) Find the time taken while accelerating.(c) Find the distance moved while accelerating.

Given that the total distance moved by the car is 852 m , find the value of T.


10

3 ms−2

2 ms−2

1.5 ms−2 12 ms−1 12 ms−1

EQUATIONS OF MOTION WITH UNIFORM ACCELERATION

• A particle is said to move with uniform acceleration if the change in velocity in any equal time interval is the same in both direction and magnitude.

• From the magnitude,

! acceleration =

! ! a =

! !! ! a =

! From (1),

! When t = 0, s = 0 , c = 0

! From (1),

! Put a into (2),

! From (1)

! Put into (2)


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Change in Velocity

Time Taken

Final Velocity - Initial Velocity

Time Taken

v - u

t

⇒ at = v − u

v = u+ at

⇒ ds

dt= u+ at

⇒ s = ut+1

2at2 + c

a =v − u

t

s = ut+1

2

�v − u

t

�t2

⇒ s = ut+1

2vt− 1

2ut

⇒ s =1

2ut+

1

2vt

t =v − u

a

s =1

2(u+ v)t ⇒ s =

1

2(u+ v)

�v − u

a

�

⇒ 2as = (u+ v)(v − u)

⇒ v2 − u2 = 2as

⇒

⇒

⇒

⇒

......(1)

......(2)

......(3)

......(4)

Example 1A car moving with a constant acceleration of passes point A with a speed of . Find the speed and the distance travelled after 8 seconds.

Example 2A particle moves with uniform acceleration. Find the acceleration if (a) the initial velocity is and the velocity after 4 seconds is (b) the initial velocity is and the velocity after 3 seconds is

Example 3A particle moves in a straight line with an initial velocity of and a retardation of . Find (a) the time taken and the displacement for the particle to come to instantaneous rest.(b) the displacement and distance travelled after 8 seconds.

Example 4At the same instant two children, who are standing 24 m apart, begin to cycle directly towards each other. Jack starts from rest at a point A, riding with a constant acceleration of and Ali rides with a constant speed of . Find how long it is before they meet.


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2 ms−1

2 ms−2

24 ms−1 4 ms−2

6 ms−1 22 ms−1

14 ms−1 8 ms−1

3 ms−2 6 ms−1

VERTICAL MOTION UNDER GRAVITY

• When an object falls under gravity, its downward acceleration is called the acceleration due to gravity and its denoted by g. To simplify circulations, we shall take the value of g to be constant at , although the value of g varies slightly at different points on the earth.

• If an object falls vertically and we neglect the effect of air resistance, its motion is a constant acceleration motion in a straight line and the equations of motion hold.

Example 1A stone is projected vertically upwards from a height of 14 m above the ground with velocity . Calculate (a) its greatest height above the ground,(b) its speed when it hits the ground

Example 2A boy throws a ball vertically upwards from a seven-metre high roof. (a) If, after 2 seconds, he catches the ball on its way down again, with what speed was it thrown?(b) What is the velocity of the ball when it is caught?(c) If the boy fails to catch the ball, with what speed will it ht the ground?


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10 ms−2

9 ms−1

Example 3A particle falls from rest from a point h m above the ground. During the last 2 seconds before it hits the ground, the particle moves through 140 m. Calculate(a) the speed at the beginning of the last 2 seconds(b) the speed at which it hits the ground(c) the total time it takes to reach the ground(d) the value of h


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distance displacement - 1634maths · pdf fileit is usually taught under the ... since this is...

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