Question in description

Question

For time t ≥ 0, the velocity of a particle moving along the x-axis is given by v(t) = x³ − 4x² + x. The initial position of the particle at time t = 0 is x = 4. Which of the following gives the total distance the particle traveled from time t = 0 to time t = 4?

Answer 1

I think it's a typo, but the velocity function is defined as "v(t)" (velocity as a function of time), but the "x" variable is used instead. Safe to say, I think the accurate function is:

v(t) = t³ − 4t² + t for time t ≥ 0

It's important to first understand the definition of "total distance." It's the entire distance traveled or covered independent of whether you switched directions and independent of relative final position (which is displacement, not total distance).

In calculus, the definition of position is: s(t) = ∫v(t)dt (integrating velocity to get position) but to circumvent the change in direction, add an absolute value around the velocity |v(t)| (which is speed-- no longer velocity since the sign (whether positive or negative) is no longer taken into account).

Total distance then is:

T.D. = ∫|v(t)|dt

= ∫|t³ − 4t² + t|dt on the interval [0,4]

From here, I recommend graphing to get the big picture:

t³ − 4t² + t = t (t² - 4t + 1) [Factor out the t]

To determine the roots (where the cubic function cross the t-axis), set the latter equal to zero.

t (t² - 4t + 1) = 0

Therefore t = 0 or t² - 4t + 1 = 0 by the zero-product property of multiplication

To solve for t² - 4t + 1 = 0, use the quadratic equation since it's not factorable (note: it's always easiest to factor if possible. In this case, it's not, so the long-route must be taken aka quadratic equation which my algebra 1 teacher appropriately called "the nuke").

t = [-b ±√(b² - 4ac)]/2a

t = [-(-4) ± √[(16 - 4(1)(1)]]/2(1)

t = (4 ± √12)/2

t = (4 ± 2√3)/2

t = 2 ± √3

The cubic function has single roots (degree 1 zeroes) at t = 0, 2 ± √3 meaning the cubic function will intersect the t-axis at those points.

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For some odd reason, there's a character limit on answers such as these (feel free to email Wyzant to correct issues like this, so I can't include my full answer). Feel free to email me at [email protected] or set up a session to send you the full answer.

Needless to say, you can numerically confirm with an online calculator or any sufficiently advanced TI calculator that the total distance the particle travels from time t = 0 to time t = 4 is 16√3 - (40/3) units or approximately 14.379 units.