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Instructions
Define the model for which acceleration will be calculated. As an example, using the displacement equation f(t) = t^3 + 4t^2 + sin(t), find the instantaneous acceleration at t = 0.5s. Recognize that while instantaneous acceleration is the derivative of instantaneous velocity, the displacement equation can be produced by taking the anti-derivative of velocity, and is key to calculating the solution.
Find the derivative of f(t) to produce an equation for the instantaneous velocity. Using the shorthand notation, d/dt [f(t)] = f'(t); t^3 goes to 3t^2, 4t^2 goes to 8t, sin(t) goes to cos(t). Therefore f'(t) = v(t) = 3t^2 + 8t + cos(t). Derive the function v(t) to produce a solution solving the instantaneous velocity, d/dt [v(t)] = v'(t). 3t^2 goes to 6t, 8t becomes a static variable of value 8, and cos(t) goes to -sin(t). The solution is v'(t) = a(t) = 6t + 8 - sin(t).
Take the equation a(t) and refer back to the defined model, which asks the instantaneous acceleration at 0.5 seconds - a(0.5) = 6(0.5) + 8 - sin(0.5) = 10.5 rounded to 3 significant figures.
Alternately instantaneous acceleration could be solved by plotting the graph f(t). With time on the x-axis and distance on the y-axis, the velocity of an object can be calculated by taking the area under the curve between two time points. From this, acceleration is simply figured out by drawing a tangent to the curve at time t = 0.5, however the result produced will not be as accurate as using derivatives but is useful for double checking your results.