How to Calculate Instantaneous Acceleration
Acceleration reflects the rate of the velocity change and is expressed in meter/(second)^2 units. Mathematically, acceleration is defined as the first derivative of the velocity. Instantaneous acceleration is the acceleration at any given moment of time. However, velocity is the first derivative of the object displacement function, and, hence, acceleration is the second derivative of the displacement. As an example, calculate the instantaneous acceleration at a time of 5.5 s if the object movement (in meters) is described with the function f(t)=t^3+5t^2-2t+14.
Other People Are Reading
Instructions
-
-
1
Consider differentiation rules that will be used in Steps 2 and 3 below.
Rule 1. The derivative of the function "t in the power of p," namely "f(t)=Ct^p," is
"df/dt=pCt^(p-1)." "C" is any constant number. A derivative is abbreviated as either "df/dt" or "f'(t)."
Rule 2. The derivative of any constant number is 0. -
2
Apply the rules from Step 1 to the function f(t) to calculate its first derivative and derive the velocity equation.
Velocity(t)=f'(t)=(t^3+5t^2-2t+14)'=3t^(3-1)+2x5t^(2-1)-1x2t^(1-1)+0=3t^2+10t-2. -
-
3
Apply the rules from Step 1 to the velocity function f'(t) (Step 2) to calculate its first derivative and derive the acceleration equation.
Velocity(t)=f'(t)=(t^3+5t^2-2t+14)'=3t^(3-1)+2x5t^(2-1)-1x2t^(1-1)+0=3t^2+10t-2.
Acceleration(t)=(Velocity(t))'=(3t^2+10t-2)'=2x3t^(2-1)+1x10 t^(1-1)+0=6t+10. -
4
Calculate the instantaneous acceleration at 5.5 s using the equation derived in Step 3. Acceleration (5.5 s)=6 x 5.5s+10=43 m/s^2.
-
1
