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