The coordinate of a particle in meters is given by x(t)= 16t - 3.0t³, where the time t is in seconds.The particle is momentarily at rest at t = A)0.75 s B)1.3 s C)1.8 s D)5.3 s E)7.3 s

Question

Quizplus · Accepted Answer

To find when the particle is momentarily at rest, we need to find when its velocity is zero. The velocity is the derivative of the position function with respect to time. The given position function is $x(t) = 16t - 3.0t^{3}$. Differentiating this with respect to time gives the velocity function: $v(t) = \frac{dx}{dt} = 16 - 9t^{2}$. Setting this equal to zero to find when the particle is at rest gives $0 = 16 - 9t^{2}$, which simplifies to $9t^{2} = 16$, and further to $t^{2} = \frac{16}{9}$. Taking the square root of both sides gives $t = \pm\frac{4}{3}$, but since time cannot be negative in this context, $t = \frac{4}{3}$ seconds or 1.3 seconds.

The Coordinate of a Particle in Meters Is Given by X(t)=