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STRAIGHT LINE MOTION.

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Presentation on theme: "STRAIGHT LINE MOTION."— Presentation transcript:

1 STRAIGHT LINE MOTION

2 Motion An object is said to be in motion, if its position changes with respect to time. This is related to the observer. If its position is not changing, the object is said to be at rest. Point object If the distance travelled by the object is very large compared to the size of the object, then the object can be considered as a point object or particle. For example, a train whose length is nearly a kilometre may be treated as a point objects if it travels very long distances. Frame of reference: It is a set of axes from which the state of body is observed with respect to time.

3 Motion in one, two and three dimensions
*If the motion of an object is restricted to a straight line it is said to execute one dimensional motion. e.g:A train running along a straight track *If the motion of an object is restricted to a plane it is said to be in two dimensional Motion. e.g.The motion of a boat on a lake *An object moving in space in said to be in three dimensional motion. e.g:A butterfly flying m air

4 Vector:- A physical quantity having both magnitude and direction
e.g:Force, velocity etc., Scalar- A physical quantity having only mangnitude e.g: mass, distance etc.,

5 Differences between the terms ‘distance’ and ‘displacement’:-
1. Distance is the length of actual path travelled by a body, irrespective of its direction of motion. 2. Distance between two given points may be the same or different for different paths chosen. 3. It is a scalar quantity. 4. Distance covered is always positive or zero. 1. Displacement is the shortest distance between the initial and final positions of a body in a given direction. 2. Displacement between two given points is always same. 3. It is a vector quantity. 4. Displacement covered may be positive, negative or zero.

6 Use the diagram to determine the resulting displacement and the distance travelled

7 Uniform speed - when an object covers equal distances in equal intervals of time, however the small these time intervals may be, it is said to be in ‘uniform speed’. Non-uniform speed - when an object covers unequal distances in equal intervals of time then it is said to be in ‘non-uniform speed’. Average speed’: When the speed of a body varies with time, we need to define its average speed.

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9 A car covers first half of the distance between two places at a speed of 40km/h and and the second half at 60 km/h. What is the average speed of the car? Velocity:- Velocity of a body is defined as the displacement produced per unit time. *It’s S.I unit is m/s.

10 Average velocity:- It is the ratio of displacement to the total time taken

11 Distinguish between the terms ‘Speed’ and ‘Velocity’
1. It is the distance travelled by a body per unit time in any direction. 2. It is a scalar quantity. 3. It is always positive or zero but never negative. 1. It is the distance travelled by a body per unit time in a given direction. 2. A vector quantity. 3. It may be positive or negative or zero.

12 Instantaneous velocity:- The velocity of particle at any instant of its motion is called instantaneous velocity. Acceleration (a):- It is the rate of change of velocity. Its S.I unit is m/s2.

13 Instantaneous acceleration:- The acceleration of an object at a given instant of time.
Uniform acceleration:- If the velocity of the particle changes equal amounts in equal interval of time, then particle is said to move with uniform acceleration.

14 Graphical representation of motion:-
*Time is always plotted on the X-axis. Distance-time graph:- It is a graph obtained by plotting distance travelled along Y-axis and time along X-axis. The uses of distance -time graph are as follows: (a) It tells the position of the body at any instant of time. (b) The distance covered by the body during a particular time interval can be seen from this graph.

15 1. The steeper the graph, the faster the motion
2. A horizontal line means the object is not changing its position - it is not moving, it is at rest 3. A downward sloping line means the object is returning to the start.

16 Velocity –time graph:- velocity along Y-axis

17 Note:- •The steeper the graph, the greater the acceleration. • Area under velocity –time graph gives acceleration Here the shaded area = 30 x 6 m2 = 180 m2 . i.e., The distance travelled in 6 second = 180m2.

18 Position –time graph Velocity –time graph Acceleration –time graph

19 Q1.Shows the velocity –time graph of a particle travelling along a straight line. Find the distance and displacement covered in 6s.

20 Equation of Motion by Graphical Method
Consider an object moving along a straight line with an initial velocity u and uniform acceleration a. suppose, it travels distance s in time t. As shown in figure its velocity-time graph is a straight line.

21 or, v - u = at  v = u + at . This proves the first equation of motion.

22 2. Equation for position-time relation:
Distance travelled by an object in time t is s, Or, s = Area of the trapezium OABE = Area of OADE + Area of ADB Or, s = (OA x OE) + (½ x DB x AD) Eq. (1) Now, DB = BE - DE = v - u = at Putting this value for DB in Eq. (1), we get s = ut + ½at  this proves the second equation of motion.

23 3. Equation for position-velocity relation:
The distance travelled by an object in time t is or, substituting EB, ED and OE with v, u and t we get, s = ½ (v + u) t Eq. (2) But from the first equation of motion we know that v = u + at

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25 Free Falling bodies The motion of a body falling under the effect of gravity alone with a constant acceleration is known as free fall. This acceleration is due to the gravitational attraction of the earth and is called acceleration due to gravity. It is represented by g. Its value is the same for all bodies at the same place. The value of g is nearly 9.8 m/s2 at our place. Note:- (i): When a body is projected vertically up , ‘g’ is taken as negative. eqns. Of motion becomes a = - g v = u - gt s = ut - ½ gt2 v2 - u2 = - 2gs (ii). When a body is projected vertically down , ‘g’ is taken as positive.  a= + g v = u + gt s = ut + ½ gt2 v2 - u2 = + 2gs

26 Q2. A body is projected vertically upwards with a velocity 49 ms
Q2. A body is projected vertically upwards with a velocity 49 ms . Find (a) the maximum height (b) time of ascend (c) time of flight (d) velocity when it is at a height of 60 m. [g = 9.8 ms-2 ] Q3. A stone is projected vertically up from the top of a tower 73.5 m with velocity 24.5 ms1. Find the time taken by the stone to reach the foot of the tower.  Q4. A stone thrown vertically up from the top of a tower wi.h . reaches the ground in 6 seconds. Find the height of the tower.


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