Equation of uniformly accelerated motion; (a) Revision on velocity-time graph (b) application and interpretation of equation of motion in simple problems.
Subject:
Physics
Term:
FIRST TERM
Week:
WEEK 4
Class:
SS 2
Topic:
Equation of uniformly accelerated motion; (a) Revision on velocity-time graph (b) application and interpretation of equation of motion in simple problems.
Previous lesson:
The pupils have previous knowledge of
Vectors; (a) concept of vectors (b) vector representation (c) addition of vectors (d) resolution of vectors(c) concept of resultant velocity using a vector representation.
that was taught as a topic in the previous lesson
Behavioural objectives:
At the end of the lesson, the learners will be able to
- Revision on velocity-time graphs.
- Derivations of equation of uniformly accelerated motion
- Application of equation of uniform motion
- Equation of motion under gravity
Instructional Materials:
- Wall charts
- Pictures
- Related Online Video
- Flash Cards
Methods of Teaching:
- Class Discussion
- Group Discussion
- Asking Questions
- Explanation
- Role Modelling
- Role Delegation
REFERENCES:
- new school Physics by MW Anyakoha
- New system PHYYSICS for senior secondary schools. Dr. Charles Chew.
- Comprehensive Certificate Physics by Olumuyiwa Awe
- Senior School Physics BY PN Okeke, SF Akande
- STAN Physics.
Content:
WEEK 4:
TOPIC: EQUATION OF UNIFORMLYACCELERATED MOTION
CONTENT
- Revision on velocity-time graphs.
- Derivations of equation of uniformly accelerated motion
- Application of equation of uniform motion
- Equation of motion under gravity
–
PERIOD ONE:
Velocity – time v-t graph.
1. Gradient of a v-t graph = acceleration
Acceleration = gradient =
t
v
2. Area under a v-t graph = distance.
v
c
e
d
b
0
a
t
i. Total distance covered during the motion = area of trapezium 0edc ii. Distance covered during acceleration = area of triangle 0ea iii. Distance covered during constant velocity = area of rectangle aedb iv. Distance covered during deceleration = area of triangle bdc v. Acceleration = slope of line 0e , vi. Deceleration = slope of dc,
Example 1
1. A car starts from rest and accelerates uniformly to 15ms-1 in 5 s. it then continues at this velocity for the next 10s before decelerating back to rest in another 8 s.
Use the information to answer the following questions
-
- Sketch the velocity time graph of the motion of the car
- Calculate the acceleration of the car
- Calculate the deceleration of the car iv. What is the total distance travelled by the car v. Estimate the average speed of the car.
v
ii.
Acceleration
iii.
deceleration
–a
=
15
23
15
0
5
t
-
- total distance = area under the graph
= area of trapezium S
=
-
- average speed v = v =
Example 2
A body at rest is given an initial uniform acceleration of 8.0ms2 for 30s after which the acceleration is reduced to 5.0ms2 for 30s. The body maintains the speed attained for 60s after which it is brought to rest in 20s.
(a) Draw the velocity-time graph of the motion using the information given above.
Using the graph, calculate (b) maximum speed attained during the motion. (c) average retardation as the body is brought to rest. (d) total distance travelled during the first 60s (e) average speed during the same intervals as in (c)
Solution.
22
|
Page
a
(
)
V(m/s)
5
ms
-2
ms
8
–
2
Stage
1
stage
0
30
s
(b)there are two stages of acceleration
Stage 1. Acelecation = gradient a= 8 ms-2
Cross multiplying
Stage 2. A= 5 ms-2
Cross multiplying
But V1=240
Average retardation is equal to gradient but V2 = 390ms-1
a = 19.5ms-2
Average retardation = – 19.5ms-2
(d) distance is in the first 60sec = area of triangle + area of the next trapezium
S =
S =
(d) average speed
RELATIVE MOTION
This is the motion of a body with respect to another. All motion is relative. The motion of a car on the road is with respect to the earth or any other frame of reference in which the motion of the car is being observed. Resultant velocity of relative motions
-
-
-
- Consider two cars X and Y travelling in the same direction and at the same speed, a commuter in X will observe that Y is stationary (not moving) Relative velocity
- If car X is to be travelling at a speed Vxwhich is greater than the speed of Vy, a commuter in car Y will observe the speed of car X to be
-
-
A commuter in X will observed the relative velocity of Y to be
This value will be negative. This means that to an observer in X, the car Y will appear to be going backward (going the opposite direction with a speed of /Vy – Vx/
But is car X and Y were to be travelling in opposite direction, the relative velocity of X with respect to Y will be
Vy – Vx = relative velocity of Y with respect to X
N.B. note that the relative velocity of X with respect to Y, Vxy is equal in magnitude but opposite in direction to the relative velocity of Y with respect to X, Vyx.
Vxy = – Vyx
EXAMPLES
-
- Two racing cars A and B travelling in the same direction at 300m/s and 340mls respectively. What is the relative velocity of A with respect to B? Solution:
Va= 300 km/h
Vb=340 km/h
Relative velocity of a with respect to B, Vab = Va -Vb = 300 – 340 = -40 km/h
(note that this is negative. A appears to be travelling in the opposite direction to B)
-
- A boat whose speed is 8 km/h sets course on a bearing 0600. If the tide is running at a speed of 3 km/h from a bearing of 3300, find;
i. The actual speed of the boat(i.e, relative speed of the boat) ii. The direction of travel
| N.B the angle 900 in the triangle is obtained by geometry |
Vb
V
t
V
b
60
0
boat
V
rel
Direction of tide
Vt
To obtain the relative velocity (actual velocity), draw the component velocity such that the head of one point to the end of the other. Draw the relative velocity to beginning from end of the first to the head of the last.
Using Pythagoras theorem
Let ϴ be the angle between the relative velocity and the direction of the boat.
EVALUATION
1. A train runs at a constant speed of 20m/s for 300s. and then accelerate uniformly to a speed of 30m/s over a period of 20s. this speed is maintained for 300s before the train is brought to rest with uniform deceleration is 30s. draw the velocity – time graph to represent the journey describe above. From the graph find,
i. The acceleration while the speed changes from 20m/s to 30m/s. ii. The total distance travelled in the time described iii. The average speed over the time described. (J.M.B)
2. A car travels at a uniform velocity of 20m/s for 4s. if the brakes were applied to bring the car to rest in the next 8 s. draw the velocity time graph for the motion. How far does the car travel after the brakes were applied?
GENERAL EVALUATION
-
-
- The planetary motions are examples of —– motion. (a) rectilinear (a) rotation (c) vibratory (d) spin
- The rate of change of velocity is called —- (a) speed (b) displacement (c) uniform velocity (d) acceleration
- Which of these is also referred to as negative acceleration? (a) instantaneous acceleration (b) uniform acceleration (c) retardation (d) non-uniform acceleration
- The gradient of a distance –time graph gives —- (a) velocity (b) acceleration (c) speed (d) displacement
- —- represent the area under a velocity – time graph. (a) distance (b) speed (c) acceleration (d) none of the above
- Which of these graphs represent the velocity-time graph of the motion of a spherical metal ball falling through a fluid until it attain it terminal velocity?
-
(
d
)
(
c
)
(
b
)
(
a
)
WEEKEND ASSIGNMENT:
1. The graph below represent the velocity time graph a body
v
e
d
b
0
a
t
Sketch the corresponding displacement –time graph for this motion.
READING ASSIGNMENT
Students should read page 121-125 of the New School Physics by MA Anyakoha and answer question 21 and 25 on page 131
DERIVATIONS OF EQUATION OF UNIFORMLYACCELERATED MOTION
Sub-Topic3: ANALYSISOF RECTILINEAR MOTION.
Supposing a body moving at an initial velocity later attains a final velocity in time Its acceleration is given as:
Change in velocity = v-u
By making v subject of the formula,
We have
Recall that
Since the body above experienced two velocities, u & v, thus, the average velocity is
Hence,
Putting (2) into (3), we have
Hence,
From (1),
Putting (5) into (3), we have
But is a different of two squares, implying that
Hence,
Making subject of the formula, we have
Thus, in summary, the equations of motion include:
Under gravity, for a body descending, . Therefore, the above equations become:
Under gravity, for a body ascending, . Therefore, the above equations become:
EVALUATION
- Define the following terms (i) average speed (ii) average velocity (iii) uniform acceleration (iv) constant velocity
- State the value of the acceleration of a body moving with uniform velocity.
PERIOD TWO
Derivation of equation of uniform motion
Recall that,
Equation (i) –Equation (ii) —
Substituting equation (i) into equation (ii)
Again from equation (i),
Dividing both sides by a,
Substituting equation (3) into equation (ii)
(ii) —
becomes
Expanding the bracket in the numerator, Cross multiplying,
Summarily, the equations of uniformly accelerating bodies are:
.N.B. note that these equations can only be applied to solve problems on bodies moving with constant/uniform acceleration. Problems on bodies moving with nonuniform acceleration can be solved using differential calculus.
EVALUATION
- State the equations of uniformly accelerating bodies.
- Derive the (iv) of uniformly accelerating motion.
PERIOD THREE
Application of the equations of uniform accelerating bodies.
- A train starts from rest and accelerate until it attains a velocity of 8m/s is 10 s.
calculate the acceleration of the train.
Solution:
For a body at rest velocity is zero.
Initial velocity U=0
Final velocity V= 8m/s
Time t=10 s
Acceleration a= ?
{ you use any of the four equations that has U,V, t, a has identified from the question}
V = U + at
8 = 0 + ax10
8 = 10a
Dividing both side by 10 a = 0.8m/s2
- A horse rider moving with constant acceleration covers the distance between two point 70.0m apart in 7.0 s. if his speed as he passes the second point is 15.0 m/s.
what is its speed at the first point?
Distance S = 70.0m
Time t = 7.0s
Initial speed U = ?
Final speed V = 15.0m/s
{ the equation containing S, t, U, and V is S = }
70 =
Cross multiplying
(15+U)7 = 140
Dividing both sides by 7
15 +U = 20
U = 20 – 15
U = 5m/s
- A body starts with an initial velocity of 26m/s and moves down it with uniform acceleration of 7m/s2 for 25 s. find the total distance moved in metres
Solution: Initial velocity U = 26m/s
Acceleration a = 7m/s2
Time t = 25 s
Distance S = ?
{the equation containing U, a, t and s is
Motion of bodies under gravity
Neglecting air resistance, motion of bodies moving under gravity (either vertical upward or downward) is an example of uniformly accelerating motion.
- A body thrown vertically upward in the earth gravitational field. When a body is thrown vertically upward from the earth surface, it retards uniformly (with acceleration of a = -g) until it attain it maximum height where its final velocity is zero. (V = 0)
If U is the initial velocity with which the body was projected vertically upward and H=S is the maximum height where it the velocity is zero (i.e, temporarily at rest before coming down) g – acceleration due to gravity
V=0 s = H a = -g is negative (retardation) where g is the acceleration due to gravity
0 = U2 + 2(-g) H
0 = U2 -2gH
2gH = U2
H is the maximum height
Again , using V= U + at
V = 0 a = -g
0 = U + (-g)t
0 = U –gt gt = U
T is the time to reach the maximum height.
If the body is thrown vertically upward and allowed to return to the point of projection, the total time of flight is given as
- Motion of a bodies falling freely under gravity
The body was initially at rest, hence the initial velocity is zero. As it falls, it velocity increase i.e it accelerate, a = g
Using ,
V2 = 0 + 2gH
This is the velocity of the body just as it it about to reach the ground
Again using
H = 0 x t + ½ g t2
H = ½ gt2
CLASSWORK; A projectile is fired vertically upward and it reach a height of 78.4 m. find the velocity of projection and the time it takes to reach the highest point.( take g = 10m/s2) Solution: initial velocity U = velocity of projection
U =?
Maximum height S =H = 78.4 m
Acceleration a = -g = -10m/s2
Final velocity V = 0 m/s ( body is temporarily at rest at the maximum height)
{ U, S, a, V}
V2 = U2 + 2(-g) H
0 = U2 – 2 x 10 x 78.4
U2 = 1568
U
U = 39.6 m/s
CLASSWORK; A body falls from a height of 80m. what is it velocity just before hitting the ground
Solution; height H = S = 80m
Initial velocity U = 0 ( body is taken be initially at rest)
Acceleration a = g = 10 m/s2 ( this is positive because the body is coming down) Final velocity V = ?
{ S, U, a, V}
A stone is dropped from a height of 196 m. neglecting air resistances; calculate the time to reach the ground.
Solution
H=S = 196 m
a = g = 10 m/s2. ( g is positive because the body is moving downward) U=0m/s ( body is taken to be initially at rest) t= ?
196 = 0xt + ½ x 10 t2.
196 = 5 t2.
EVALUATION:
- A stone was thrown vertically upward with an initial speed U. If g is the acceleration of free fall, show that the time taken for the ball to return to its point of projection is
- A ball is thrown vertically upward with a velocity of 19.6m/s. what distance does it travels before coming to rest momentary at the maximum height?
- With what velocity must a ball be projected vertically upward for it return to it point of projection in 5s?
- A vehicle which starts from rest is accelerated uniformly at the rate of 5m/s2 for 5 s. It attains a speed which is maintained for 60 s. the vehicle is then brought to rest by a uniform retardation after another 3 s. determine the total distance covered.
GENERAL EVALUATION
- Two cars A and B move parallel to each other but in opposite direction. If the velocity of A is 10m/s and that of B is 15m/s. what is the relative velocity of B with respect to A?
- A car travelling with a uniform acceleration of 3m/s2 starts from rest. What time will it attain a velocity of 15m/s?
- A ball was thrown vertically upward from the ground with a velocity of 40m/s. a similar ball was thrown 1 s later from the same spot with the same velocity. At what time will the two ball meet each other?
WEEKEND ASSIGNMENT
Derive an equation of uniformly acceleration motion that involves only S, V, t and a
READING ASSIGNEMENT
Students should read pages 125, 130-132 of New School Physics. By MA Anyakoha and answer question26 on page 131
