Suppose you know the position p of an object as a function of time t. We'll write the position as p(t) to denote this relationship. You should be aware that the velocity of the object is given by the derivative of position with respect to time, so we'd write v(t) = p'(t). Likewise, the acceleration of the object, a(t), is the derivative of the velocity, so a(t) = v'(t) = p''(t), the second derivative of the position. Also, you should know how to use integrals to convert velocities into positions, and accelerations into either of the other functions.
Most problems you've encountered give you one of p(t), v(t), or a(t), and some other information, and you're supposed to find the other two functions. Here, we're only going to tell you some scattered facts, and you need to determine the acceleration.
Here is a simple case. Suppose you know that a(t) = K, where K is a constant and v(0) = 0. If you want v(10) = 100, what should K be? To answer this, we know that a(t) = v'(t), so v(t) must be Kt + C, where C is a constant. But, 0 = v(0) = K 0 + C = C. So, 100 = v(10) = 10 K, and so K = 10.
Now, let's complicate matters a bit. Suppose we want a linear acceleration a(t) such that v(0) = 0, v(10) = 10, p(0) = 0, and p(10) = 100. How can we find such a linear function a(t)? First off, since a(t) is linear, we need to find constants A and B such that a(t) = At + B. Integrating a(t) to get v(t), we find
v(t) = At²/2 + Bt + C.Using v(0) = 0, we find C = 0. Integrating v(t) = At²/2 + Bt to get p(t), we find
p(t) = At³/6 + Bt²/2 + D.Using p(0) = 0, we find D = 0. Therefore,
v(t) = At²/2 + Bt and p(t) = At³/6 + Bt²/2.Now, use v(10) = 10, and p(10) = 100. We find that
10 = 50A + 10B and 100 = 1000A/6 + 50B.Solving this system of equations gives us
A = -0.6 and B = 4.0and so our linear acceleration is
a(t) = -0.6 t + 4.0
The steps in doing the problems in the following crash test simulation are the same as those used above.