Motion In Straight Line

| Motion is the change of position of an object from one place to another. There are two types of |
| motions; i.Circular motion-Is the motion of an object in a circle. Examples; a/. motion of the |
| electron around the nucleus of an atom b/.r evolutionary movement of the earth around the sun. |
| ii.Linear motion-Is the motion of an object in a straight line. |
| Distance and Displacement |
| Difference between Distance and Displacement |
| Distinguish between Distance and Displacement |
| Distance Is the length between two points or two objects.It is a scalar quantity i.e. it has a |
| magnitude only but no direction. Its symbol is X or S and the SI unit is meter (m). Other units |
| used are Kilometer (km) and Centimeter (cm) |
| Displacement Is the distance in a specific direction. It is a vector quantity i.e. it has both |
| Differences between distance and displacement |
| Is the length moved by an object between two points Is the distance in a specific direction |
| The SI Units of Distance and Displacement |
| State the SI units of Distance and Displacement |
| The symbol for distance is X or S and the SI unit is meter (m). Other units used are Kilometer |
| The standard unit of displacement in the International System of Units ( SI ) is the |
| Difference between Speed and Velocity |
| Distinguish between Speed and Velocity |
| Speed is the distance moved by an object in a unit time or is the rate of change of distance . It is a |
| Velocity is the displacement moved by an object per unit time or is the rate of change of |
| displacement. It is a vector quantity and its symbol is U or V. |
| Formula for speed and velocity |
| Speed(velocity) = Distance/displacement x time taken |
| Differences between speed and velocity |
| Is the rate of change of distance Is the rate of change of displacement |
| It is a scalar quantity Is a vector quantity |
| The SI Unit of Speed and Velocity |
| State the SI unit of Speed and Velocity |
| The SI unit of speed and velocity is meter per second (m/s). Other units are Km/h or cm/s |
| The Average Velocity of a Body |
| Determine average velocity of a body |
| – Is the velocity of a body at the start of observation |
| – is the velocity of a body at the end of observation |
| –is the average or mean between initial and final velocity or is the ratio |
| of the total displacement to the total time. |
| Uniform or constant velocity |
| Is the one whereby the rate of change of displacement |
| is the actual velocity of a moving object recorded by a stationary |
| is the velocity of a moving object recorded by a moving observer . |
| is the velocity of a moving object recorded at any time. |
| Acceleration is the rate of change of velocity or is the change in velocity per unit time |
| Mathematically. Acceleration, a = (final velocity, v – initial velocity, u)/time, t |
| Interpret velocity time-graph |
| This is velocity against time graph. Consider a body accelerating uniformly from rest to a certain |
| velocity v within time t. This can be represented graphically as shown below; |
| The distance x moved by the body is given by the area under the curve. |
| In this case is the area of triangle OBC. |
| The acceleration is given by the slope of the triangle OBC. |
| The Acceleration of a Body |
| Determine the acceleration of a body |
| A car starts from rest and accelerates to a velocity of 120m/s in one minute. It then moves with |
| this speed for 40seconds finally decelerates to rest after another 2 minutes. Calculate; |
| the total time taken for the whole motion |
| 1st stage; acceleration u =0, v = 120m/s,t1 =1min=60s |
| 2nd stage; uniform vel u=v=120m/s,t2=40s |
| 3rd stage; Deceleration u=120m/s,v=0,t3=2min=120s |
| 1st stage; s =average vel x time = ( (120+0)/2)60 = 3600m |
| 2nd stage; s = vt = 120 x 40 = 4800m |
| 3rd stage; s = average vel x time = ((v=u)/2)t = ((120+0)/2)120 = 7200m |
| Total distance, s T =3600+4800+7200 =15600m |
| Total time taken = 60 + 40 + 120 = 220s |
| Average velocity = total distance/time |
| The Concept of Retardation |
| Explain the concept of retardation |
| Deceleration (retardation) is the rate of decrease of velocity or is the decrease in velocity per unit |
| time. Uniform acceleration or retardation Is the one whereby the rate of increase or decrease of |
| velocity is constant or it doesn’t change. |
| when a body starts f rom rest or is brought to rest, its velocity is zero |
| when the velocity of a body is constant or uniform, its acceleration is zero |
| when the velocity of a moving object increases, its acceleration becomes positive |
| when the velocity of a moving object decr eases, its acceleration becomes negative called |
| Equations of Uniformly Accelerated Motion |
| Equation of Uniformly Accelerated Motion |
| Derive equation of uniformly accelerated motion |
| There are three equations of motion; |
| Newton’s first equation of motion |
| It is given by; v = u + at |
| From the formula of acceleration; a = (v-u)/t |
| Newton’s second equation of motion |
| Suppose an object starts from rest with intial vel,u to final vel,v after time t The distance ,S |
| moved is given by S = Average vel x time S =( (u+v)/2)t |
| Substitute Newtons first eqn,gives; S = ((u+u+at)/2)t |
| Newtons third equation of motion |
| where v=final velocity; u =initial velocity; a = acceleration; s = distance covered. |
| From Newton’s first eqn; v = u + at |
| Equations of Accelerated Motion in Daily Life |
| Apply equations of accelerated motion in daily life |
| Apply equations of accelerated motion in daily life |
| The Concept of Gravitational Force |
| Explain the concept of gravitational force |
| The acceleration of a free falling body is known as the acceleration due to gravity denoted by g |
| and controlled gy gravitational force of the earth. |
| When two bodies of different masses are released from a certain height h above the ground, they |
| will reach the ground at different times with the heavier one reaching earlier before the lighter |
| The reason is that the air resistance is more effective on lighter bodies than in heavier bodies, |
| consequently this affect acceleration due to gravity in a reverse manner. |
| But dropping a light object and a heavy object in a vacuum they will reach the ground at the |
| same time due to the absence of air resistance effect. |
| Consider the following two cases; |
| Consider a body falling freely from a certain height h and uses time t to reach the ground. |
| In this case: acceleration, a = acceleration due to gravity g |
| Consider a body thrown vertically upwards from the ground with an initial velocity u to a certain |
| height h and then comes back to the ground after time t. |
| Final velocity at the ground = v |
| Acceleration due to Gravity |
| Determine acceleration due to gravity |
| A simple pendulum is a small heavy body suspended by a light inextensible string from a fixed |
| support and it is normally used to determine acceleration due to gravity. |
| It is made by attaching a a long thread to a spherical ball called a pendulum bob. The string is |
| held at a fixed at a fixed support like two pieces of wood held by a clamp and stand. |
| If the bob is slightly displaced to position B,it swings to and from going to C through O and back |
| to B through O. When the pendulum complete one cycle/revolution the time taken is called the |
| period of oscillation, T. |
| is the time taken by the pendulum bob to complete one complete cycle. |
| is the angle made between the string and the vertical axis when |
| the bob is displaced to a maximum displacement. |
| is the maximum displacement by which the pendulum has been displaced. |
| of pendulum is the length of the string from the point of attachment on the |
| wooden pieces to the canterof gravity of the bob. |
| From the experiments, it has been observed that, changing the weight of the bob and keeping the |
| same length of pendulum, the period is always constant provided that all swings are small though |
| they may be different in size. |
| The period T of the pendulum is given by; |
| Where; l = length of pendulum; g = acceleration due to gravity |
| It follows that if we plot a graph of l against T. it is going to be a straight line with a slope |
| and y –intercept equal to 0. |
| When the bob is raised to point B it will gain potential energy and the bob will swing due to the |
| conservation of energy from potential to kinetic energy. |
| At B and C all energy is P.E. |
| If the pendulum swung in vacuum the oscillations would have been continuous. But practically, |
| air friction causes losses of energy of the pendulum bob. That is why after a certain time the |
| The Application of Gravitational Force |
| Explain the application of gravitational force |
| Determination of acceleration due to gravity by using the simple pendulum. |
| A simple pendulum, stand, clamps and stop watch. |
| Tie a piece of thread to a brass bob(about 100g) |
| Suspend a bob with a thread between wooden pieces by clamping them on to a stand. |
| Pull the bob slightly to one side and release the bob. Make sure the bob swings 50 |
| To be more accurate, perform three measurements for each length l of the pendulum. |
| Repeat the procedures with l = 60cm and 50cm. |
| Record the results as in the table below. -Plot the graph of l against T |
| Length, l (cm) Time for 50 oscillations. T (s) T² (s²) |
| The graph obtained will be a straight line through the origin. |
| The slope of the graph is given by; |
| From this equation, the acceleration due to gravity can be computed easily. |
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