## 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 bikerâ€™s acceleration when t = 2.

##
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__**Hint**

**Hint**

Velocity is a derivative of distance.

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__**Hint 2**

**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