Solving Acceleration Problems

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In other words an object whose velocity is v has an acceleration of which is read as "dv - dt", and is the rate of change of the velocity v, with respect to time.

The big advantage of using calculus to solve speed and acceleration problems is that the velocity can be written as an expression, not just a fixed number, so very complex problems can be solved where the acceleration is not constant.

So a unit for force is actually the kilogram-meter per second squared. The unit for force is named after Isaac Newton, and it is called the ' Newton', abbreviated ' N'.

One Newton is one kilogram-meter per second squared.

Position, velocity, and acceleration problems can be solved by solving differential equations. S(0) equals 0 means I plug in 0 for position when time is 0.

Acceleration is the derivative of velocity, and velocity is the derivative of position. Now the integral of t is 1/2t² so k times t is going to be k over 2, t² plus a constant. So 0 for position equals, k over 2 times 0² plus D, that tells me that D equals 0.

Kinetics is the study of forces acting on these bodies and how it affects their motion.

--------------------------- Recommended Background: To be successful in the course you will need to have mastered basic engineering mechanics concepts and to have successfully completed my courses en titled an “Introduction to Engineering Mechanics” and “Applications in Engineering Mechanics.” We will apply many of the engineering fundamentals learned in those classes and you will need those skills before attempting this course.

So, to find the position function of an object given the acceleration function, you'll need to solve two differential equations and be given two initial conditions, velocity and position.

Since a(t)=v'(t), find v(t) by integrating a(t) with respect to t. That tells me that my velocity function is actually k times t. Now let’s take this and write it as a differential equation because v(t) is the same as s'(t).

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