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A particle moves along the
x-axis so that at time
t >= 0 its velocity is given by
v(t)=3t^(2)+12 t-36. Determine all values of
t when the particle is at rest.
Answer:
t=

A particle moves along the $x$ -axis so that at time $t \geq 0$ its velocity is given by $v(t)=3 t^{2}+12 t-36$ . Determine all values of $t$ when the particle is at rest. $\newline$ Answer: $t=$

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. A particle moves along the

x

-axis so that at time

t \geq 0

its velocity is given by

v(t)=3 t^{2}+12 t-36

. Determine all values of

t

when the particle is at rest.

\newline

Answer:

t=

Set Velocity Function Equal: To find when the particle is at rest, we need to set the velocity function $v(t)$ equal to zero and solve for $t$ . $v(t) = 3t^2 + 12t - 36$ $0 = 3t^2 + 12t - 36$
Simplify by Dividing: We can simplify the equation by dividing all terms by $3$ , which is the greatest common divisor of the coefficients. $\newline$ $0 = t^2 + 4t - 12$
Factor Quadratic Equation: Now we need to factor the quadratic equation $t^2 + 4t - 12$ . $(t + 6)(t - 2) = 0$
Solve for Values of t: Setting each factor equal to zero gives us the values of t when the particle is at rest. $\newline$ $t + 6 = 0$ or $t - 2 = 0$
Consider Domain: Solving each equation for $t$ gives us the two values when the particle is at rest. $t = -6$ or $t = 2$
Final Value of t: However, we must consider the domain of the problem, which is $t \geq 0$ . Since $t = -6$ is not within the domain, we discard it. $\newline$ The only value of $t$ when the particle is at rest is $t = 2$ .

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