How do you derive the equation for position vs. time from the velocity vs. time equation? For example let's say the velocity equation is v = 0.17t + -0.43
How would I go about figuring out the position vs. time equation with accurate numerical constants?
Thanks in advance!
To derive the equation for position vs. time from the velocity vs. time equation, you need to integrate the velocity equation with respect to time. This will give you the equation for position.
In your example, the velocity equation is given as v = 0.17t - 0.43.
To integrate this equation, you need to find the antiderivative of the velocity function. The antiderivative represents the indefinite integral of the function, which essentially reverses the differentiation process.
For the velocity function v = 0.17t - 0.43, the antiderivative (or the integral) with respect to time is:
∫v dt = ∫(0.17t - 0.43) dt
To integrate, you need to apply the power rule of integration. When you integrate a monomial function, you add 1 to the power and divide by the new power. The antiderivative of t^n is (t^(n+1))/(n+1).
Using this rule, the integral becomes:
∫v dt = (0.17/2)t^2 - 0.43t + C
Where C is the constant of integration. It represents the family of curves that result from integration, and it can be determined by specifying an initial condition or boundary condition.
In this case, since we are looking for the position equation, we will assume that at t = 0, the position is also 0. This means that when t = 0, the constant of integration C is 0.
Therefore, the position equation can be obtained by integrating the velocity equation and setting the constant of integration to 0:
s(t) = (0.17/2)t^2 - 0.43t + 0
Thus, the equation for position vs. time in this example is s(t) = (0.17/2)t^2 - 0.43t.