 Hello, everyone. I am Swati Ghadge, Assistant Professor, Department of Civil Engineering from Vulture Institute of Technology, Soolapur. Topic for today's session is linear motion. Learning outcome of this session. At the end of this session, learner will be able to determine displacement, velocity, acceleration and time of any object which is in linear motion. Let us see first, what is mean by displacement and distance. Distance, it is a scalar quantity since it has only magnitude. But the displacement, it is defined as the linear distance between two positions of body in the beginning and at the end of time interval. Since displacement has the magnitude and direction, it is a vector quantity. Unit of displacement is meter. Velocity and speed. The rate of change of distance with respect to time is speed and the rate of change of displacement with respect to time is called as velocity. Velocity is ds by dt. Unit of velocity is meter per second. Acceleration and retardation. The rate of change of velocity with respect to time is called as acceleration. The positive acceleration is simply acceleration and the negative acceleration is called as retardation. Acceleration is dv by dt and the unit is meter per second square. Let us see now displacement time curve. For that, take displacement on y axis. Take a time on x axis. The red color line is a displacement time curve. Draw a tangent at any point and the slope of the tangent is theta. Theta is equal to vertical by adjacent that is ds by dt and it is velocity. So the slope of displacement time curve gives velocity. Now we will see the velocity time curve. Take velocity on y axis, time on x axis. At time interval t1, say velocity is v1 and at time interval t2, say velocity is v2. So for the small interval of time say dt, the velocity is dv and at any point of the curve draw a tangent and the slope of the tangent is theta is dv by dt and it is acceleration. So the slope of the curve represents acceleration and the area under curve is v into t and that v into t is nothing but the displacement. So from this curve, we get acceleration that is the slope of the curve and displacement which is the area under curve. Acceleration time curve. Acceleration time curve gives velocity which is the area under curve that is a into t is velocity. Now you pause a video here and list out all the formulae for displacement, velocity, acceleration and time that we learn from the st curve that is displacement time curve, velocity time curve and acceleration time curve. These are the formulae that we learn from this 3 curve. First formula is velocity is equal to distance upon time that is v is equal to ds by dt that is we learn from st curve. A is equal to dv by dt that is the slope of vt curve is acceleration. So from vt curve we learn the formulae. It can be written in another way that is d2s by dt square it is the second derivation of displacement with respect to time. It is also acceleration and the area under vt curve that is v into dt is nothing but the ds or s is equal to integration of vt and from the at curve we learn the formula dv is equal to a into dt or simply v is equal to integration of a into dt. So these are the basic formulae that we learn from the graph. Now we will derive some equations for the uniform acceleration. So from the vt curve let us understand this. Initially the velocity is u and final velocity say v and the time required to change the velocity from initial velocity to final velocity is t. The slope of that line say theta, theta is equal to v minus u that is the vertical distance of this graph and horizontal distance is t. So slope is equal to v minus u upon t and we have learned that the velocity time curve gives acceleration when we plot the slope of the line. So acceleration is v minus u upon t where v is the final velocity u is the initial velocity and t is the time taken for change of velocity from u to v. Now we will derive some well known equation of linear motion. So we will continue with the equation a is equal to v minus u upon t. The same equation can be written in the another form that t is equal to v minus u upon a. It is simply the cross multiplication. From that we can write the equation v is equal to u plus at. It is a very common equation generally used to solve a numerical. Now as we know that the displacement is given by the average velocity into time. So s is equal to average velocity means v plus u divided by 2 into time. Now in that equation put v is equal to u plus at what we will get s is equal to u plus u plus at divided by 2 into t. So this is the second common equation that is s is equal to ut plus half of at square. They all are derived from the same equation that is a is equal to v minus u upon t. Now we will derive the third equation. We will continue from the displacement is average velocity into time. Put t of t is equal to v minus u upon a. Put in the equation after putting we get s is equal to v square minus u square upon 2a. Or simply we can write the same equation in another form v square minus u square is equal to twice a s. So all the equations I highlighted in a red color are the commonly used in the linear motion equation to solve the numerical. So please remember that equation with the help of that equation now we will solve that numerical. Question is a small steel ball is short vertically upward from the top of the building 25 meter above the ground with an initial velocity of 18 meter per second. In what time it will reach the maximum height? So let us first understand what is given. Height of the building is given 25 meter and from the top of the building the steel ball is thrown vertically upward with the initial velocity 18 meter per second. And after covering some distance its velocity will become 0 and then it will try to move down. Our question is at what time it will reach at this point where the velocity is 0? So let us first sort out the given data. What is given? Initial velocity is given 18 per meter second. It is we are talking for particularly upward motion. When it is moving vertically upward in that case initial velocity is given 18 meter per second. Final velocity is given 0. Acceleration is minus 9.81 meter per second square minus 9.81 because it is moving against gravity and the displacement is shown in the figure as h. So take s is equal to h and the time taken to reach maximum height take it as a t1 and we have the three equations that we have and we have three equations that we have derived. v is equal to u plus at, s is equal to ut plus half at square, v square minus u square is equal to twice as. Among these three you just check what is given data available with us and what equation is useful? So I think first equation is useful because we have final velocity, initial velocity and acceleration and we have to find out the t. So this is a useful equation in this case. So we will use the first equation. Put v is equal to 0, initial velocity is 18, acceleration is minus 9.81, t is t1 and find out t1. So we will get t1 is 1.83 second means the steel ball will take 1.83 second to reach a maximum height and at that height its velocity will become 0 and then it will start moving down with the help of gravity. So like this we have calculated the time required for the steel ball to reach at the maximum height. These are my references for this video. Thank you very much for watching.