## Acceleration

At one section of a road, a biker travels in a straight line so that his distance (D) from a point on the pavement after time (t) is D = 3t^4 - 2t^2. Determine the biker’s acceleration when t = 2.

Hint
Velocity is a derivative of distance.
Hint 2
Acceleration is a derivative of velocity.
Velocity is a derivative of distance.
$$v=\frac{dr}{dt}$$$where $$v$$ is the instantaneous velocity, $$t$$ is time, and $$r$$ is position. The power rule for the first derivative: $$\frac{d}{dx}[x^n]=n\cdot x^{n-1}$$$
Thus,
$$D=3t^4-2t^2$$$$$v=(4)3t^{4-1}-(2)2t^{2-1}=12t^3-4t^1$$$
Acceleration is a derivative of velocity.
$$a=\frac{dv}{dt}$$$where $$a$$ is instantaneous acceleration, $$t$$ is time, and $$v$$ is instantaneous velocity. Applying the power rule for the first derivative to the velocity function: $$a=12(3)t^{3-1}-(1)4t^{1-1}=36t^2-4$$$
Next, substitute $$t=2$$ :
$$a=36(2)^2-4=144-4=140$$\$
140