[Music] welcome back in our last video we talked about position velocity and acceleration so let's just summarize the equations we came up uh in those units so first of all we defined position as our variable s we said that our instantaneous velocity was V and this was equal to our first derivative of our position with respect to time and then we also defined a as our instantaneous acceleration and a was equal to thetive of our velocity so we have two equations of most that relate velocity to position and acceleration to Velocity and if we rearrange these and substitute them together we can get a third equation which relates acceleration to Velocity and position so we don't have a function in terms of time anymore so we could also write a third equation that is acceleration is equal to our velocity times the derivative of velocity with respect to position and if we rearrange this we can get it into a different form acceleration times the increment in position is equal to Velocity times the increment in velocity so if we have position and we want to go to Velocity we have to differentiate and if we have velocity and we want to go to acceleration we have to differentiate again likewise we can go in the Direction and if we have acceleration we want to go to Velocity we want to go to position we can integrate so we can write position going to Velocity going to acceleration acceleration we need to differentiate but if we want to go the other way and we want to go to Velocity or if we're at velocity and we want to go to position we need to integrate those relationships so let's rewrite these now in our integral forms the first equation if we take our acceleration and velocity equation so let's label these let's take equation let's call it equation one 2 and three so let's start with equation one then so equation one in the integral form if we rearrange rearrange this we would have velocity time DT is equal to DS and if we integrate both sides and we solve for that DS term we can write the integral from s0 to S DS is equal to our velocity or the integral from V 0 to V of our velocity time DT and if we do the same process for equation two we can say that the integral from V 0 to V of DV is equal to the integral from 0 to sometime T of our acceleration time DT and our last equation we take the integral of both sides on the left hand side we would have the integral from s0 to S of a d s and that's going to be equal to the integral V 0 to V of V VV so these are our three equations of motion for a particle in one dimension both in the differential and integral forms and using these three diff three equations or six equations we can solve almost any problem of onedimensional particle motion so there's one simplification that we like to do with these equations and that is assuming that acceleration is constant there are a lot of problems where the acceleration of motion uh is under constant acceleration and that's because a lot of particle motion or object motion is acting under the force of gravity and gravity is uh a constant acceleration force acting on everything so we can simplify the above integration equations if we have acceleration as constant so we would call our constant acceleration now we would just say our a term is some const so if we're on Earth we would say our acceleration our constant acceleration would be equal to G which is acceleration due to gravity and that's 9.81 me per second squared or we could also write this as 32.2 feet per second squ so let's look at our first equation our equation for position s0 to S DS is equal to integral from 0 to T of V DT well we need to First figure out what is our equation for uh velocity so maybe we need to do another step first and we will look at another equation and if we can figure out what that equation is for velocity we can substitute that in here to actually do this integral so let's skip ahead right now to equation two equation two is the integral from v0 of DV which equals the integral from Z of our now constant acceleration term times DT so if the acceleration term is constant we can pull that outside of the integral and we get V minus V 0 is equal to AC time T we can rearrange this into a different form where you could write B is equal to B 0 plus constant acceleration time time which is great now we have an equation for our velocity that we can put into this equation here and if we do that we can rewrite the right hand side as the integral from 0 to T of V 0 + a DT so if we rearrange this whole equation we'll end up with the equation s is equal to s0 + V 0 * t + 1 constant acceleration term time T and our last equation equation three was the integral from V 0 to V of V * DV which is equal to the integral of s0 to S our constant acceleration terms times DS so if we solve this we substitute in our equation for velocity we will get v^2 is equal to V 0^ 2 + 2 * our constant acceleration time our change in position s minus s0 so these are our three equations 1 2 3 three are three equations of motion for constant acceleration in one dimension so we call these are let do this in a different color are onedimensional equations of motion for constant acceleration so it's very important that we remember that these equations are only valid when we have constant acceleration if acceleration is changing with respect to time then we can't integrate equation two like we did and use that expression for velocity in the other two equ or other two expressions and our integrals would be different so we would have to use the general formance of our equations for the condition where acceleration isn't constant Let's do an example now in this example we have a particle traveling along a straight line to the right with a velocity V is equal to 4T minus 3t^2 m/s where T is in seconds it also tells us that s is equal to0 when T is equal to Z and the question is asking us to find the position and acceleration of of the particle when T is equal to 4 seconds so we're given a function for velocity we're given some initial conditions s0 and T is equal to Z and it's asking us to find position and acceleration when the particle of the particle when T is equal to 4 seconds so after 4 seconds so we can start by finding the acceleration and we can write the acceleration from our equations before is equal to the derivative of velocity with respect to time so the of 4T - 3T ^2 which is equal to T or Sorry 4 minus 6 T and it's asking us to find the acceleration at a time T is equal to 4 seconds so if we substitute 4 seconds into here so at 4 seconds or let's just write when T is equal to 4 we get acceleration is equal to 4 - 6 * 4 which is - 20 so next we need to find the position of the particle at T is equal to 4 seconds so to do this we can use our integral form of our equation and notice we can't actually use our constant acceleration because we just found out that acceleration is a function of time right here so it is not constant so we have to use our general form equations and we have to do the integral of this this equation for velocity so the equation we can use is the equation that relates position to Velocity and that is the integral from S 0 to S DS is equal to the integral 0 to T of our velocity function times DT so we can write this as s0 is just zero so the integral on the left hand side is just s and that is equal to the integral from 0 to T of our velocity function 4T minus 3 T ^ 2 DT and this is equal to 4 T ^2 / 2 - 3 T cubed over 3 all evaluated from 0 to T and we care about when T is equal to 4 seconds so valuate from 0 to 4 and if we do this we get our answer to be - 32 M so our position after time is equal to uh 4 seconds is 32 m in the negative Direction