Acceleration

#394 / Dynamics Math Medium

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.

Expand 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


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