 Hello friends, Myself Sandeep Javeri, Assistant Professor of Civil Engineering from Valchand Institute of Technology Sholapur. In today's session we are going to discuss the problem on linear motion for horizontal type motion. The learning outcome at the end of this session students will be able to solve problem on linear motion. Now linear motion in this the body is in motion in a straight line it is known as linear motion. For example a car is moving on the road and a body is projected vertically upward in the air with a certain velocity. Another example is a body is falling vertically downwards. So if car is moving on a road so that motion is in horizontal plane so that is called horizontal motion. Now body is projected vertically upwards or downwards so it is a vertical motion. The linear motion how may be uniform linear motion with constant velocity or 0 acceleration since the slope is 0 we have dv by dt is equal to 0 that means acceleration is 0 and non-uniform linear motion with variable velocity or non-zero acceleration. So acceleration is variable here the third case is uniform acceleration. The linear motion may have the motion in which the body have uniform acceleration. Now let us consider the equations of motion with uniform acceleration there are three equations of motion. The first one is v is equal to u plus at second is s equals to u into 2 u into t plus half at square the third equation is v square is equal to u square plus 2 as where s is a displacement u is a initial velocity and v is a final velocity and t is a duration of a body in a particular motion. Now let us consider an example a problem on horizontal motion with uniform acceleration. Three marks a b and c are at a distance of 100 meter each are made along a straight road a car starting from rest and with uniform acceleration passes the mark a and takes 10 seconds to reach b and further 8 seconds to reach the mark c. Calculate the magnitude of the acceleration of the car the velocity of the car at a the velocity of car at b and the distance of the mark a from the starting point. So here we have to find out the four things let us find out the solution for this given problem find out the solution of the given problem. Let us consider this is a point O from where the car is started to move along a straight path. So let us consider this is a point a this is a mark b this is a mark c the distance between a to b and b to c is same that is 100 meter since the car is moving from rest at O the initial velocity here it is 0 then car passing from a then a to b then b to c that velocity at a is b a velocity at b is b b here we have to find out the velocity of a and velocity of b and also we have to find out what is the distance of the car from rest up to the mark a. So let us consider the distance is x so given the time taken by the car to move from a to b is say t 1 which is equals to time taken by car from a to b which is given as 10 second let t 2 time taken by car from b to c which is equals to 8 second let a be the acceleration of car during its motion. So let us consider motion of car from a to b using equation of horizontal motion that is s equals to ut plus half at square so here displacement is given as a to b is 100 meter so 100 equals to initial velocity of car is u a so this is initial velocity of car or velocity of car at a is u a and the time from a to b is 10 second plus half of acceleration into time is 10 bracket square so this is equation number 1 further we solve this 100 is equal to 10 u a plus 50 a this is equation number 1 so this is the equation number 1 now let us consider motion of car from a to c again using equation s equals to ut plus half at square here displacement is 200 meter and initial velocity is u a and time taken is from a to c is 18 second plus half into a into 18 square we are getting 200 is equal to 18 u a plus half into a into 18 square so by simplifying this we are getting 200 is equal to 18 u a plus 160 to a so this is equation number 2 so solving equation 1 and 2 we get acceleration is equal to 0.278 meter per second square and velocity of car at a is equal to 8.61 meter per second now let us consider motion of car from b to c so we have from b to c the distance is 100 meter velocity is u b so we have using s equals to ut plus half at square put the value of s as 100 meter that is equals to velocity of car is u b and time taken is 8 second plus half into the acceleration is 0.278 times t is 8 second so bracket square so by solving this velocity of car at b is equal to 11.38 meter per second so velocity of car at a is equal to 8.61 meter per second and velocity of car at b is equal to 11.38 meter per second acceleration of car is equal to that is a is equal to 0.278 meter per second square now let us consider distance of mark a from starting point o so consider motion o to a so we have this is o this is a so the distance is x so initial velocity of car is 0 because it is starting from rest so using v square is equal to u square plus 2 as we can find the distance so here v is nothing but velocity of car at a which is 8.61 meter per second 8.61 square is equal to initial velocity of car is 0 plus 2 times 0.278 times x so value of x will be 133.33 meter so we are getting the distance of car from rest that is from o to a is 133.33 meter to pause the video and answer this question so this is a solution car is moving with the velocity of 5 meter per second after 10 second it attains final velocity of 15 meter per second so of the car can be obtained by using the expression v minus u divided by t so final velocity is 15 initial velocity is 5 and time is 10 second so after putting this we are getting the accession of car is equal to 15 minus 5 divided by 10 that is equals to 10 by 10 that is equals to 1 meter per second square these are the references that we are using for creation of this video thank you