How to Find Acceleration with Calculus

> Find Acceleration

Image:Morio|Wikimedia Commons

One of the applications of calculus is in the assessment of rates of change in physics. Acceleration is a measure of the rate in change of the velocity of an object. This means if you know the velocity of an object and how it changes, you can find acceleration by performing differentiation on the function.

Find Acceleration: Steps

Note: You can find acceleration when given either a function of the distance an object travels over time, or a function of an object’s velocity over time. If you are given the velocity instead, skip ahead to step 3.

Step 1: Given a function for the distance an object travels over time, set the derivative of the function with respect to time as equal to the velocity of the object. We will use a simple equation for this example.
x(t)=t2+2t+4
V(t) = d/dt t2+2t+4
Step 2: Solve for velocity. There are a variety of ways to do so, but here we will use three calculus rules:
f`(xn)=nxn-1
f`(K)=0 where K is a constant,
f`(x+y)=f`(x)+f`(y)
t2 becomes 2t by the f`(xn) rule, also known as the Power Rule.
2t qualifies for f`(xn) since 2t is equivalent to 2t1. By the Power Rule it becomes 1*2*t0. Since t0 is equal to 1, 2t becomes 2.
Finally, the 4 is removed since the derivative of a constant is always 0.
This will give you:
V(t)= d/dt t2+2t+4 = 2t+2
Step 3: Set the derivative of the function for velocity with respect to time equal to the object’s acceleration. It is worth noting that this is equivalent to the second derivative of the first function for displacement of the object.
d2x/dt2 = d/dt 2t+2 = a(t)
Step 4: Solve for the derivative of the velocity with respect to time. The same rules listed in step 2 apply to step 4.
dV/dt = d/dt 2t+2 = 2 = a(t)
This means the acceleration of the object has a constant value of 2.

That’s how to find acceleration!
Tip: When there are leftover variables in the formula for acceleration, this means the acceleration is not constant over time. Be sure to include any units of measure in your answer to make it correct.