Question

a particle moves along the x - axis so that its position at time t > 0 is given by x(t) and $\frac{dx}{dt}=-10t^{4}+9t^{2}+8t$. the acceleration of the particle is zero when t =
a 0.387
b 0.831
c 1.243
d 1.647
e - 0.094

Explanation:

1. First, recall the relationships between position, velocity, and acceleration:

• The velocity \(v(t)\) of a particle is the derivative of the position - function \(x(t)\), i.e., \(v(t)=\frac{dx}{dt}\). Here, \(v(t)= - 10t^{4}+9t^{2}+8t\).
• The acceleration \(a(t)\) of a particle is the derivative of the velocity - function, so \(a(t)=v^\prime(t)\).
• Differentiate \(v(t)= - 10t^{4}+9t^{2}+8t\) with respect to \(t\) using the power rule \(\frac{d}{dt}(t^{n})=nt^{n - 1}\).
• \(a(t)=\frac{d}{dt}(-10t^{4}+9t^{2}+8t)=-40t^{3}+18t + 8\).

1. Then, set the acceleration equal to zero:

• We need to solve the equation \(-40t^{3}+18t + 8 = 0\). Divide the entire equation by \(2\) to simplify: \(-20t^{3}+9t + 4 = 0\), or \(20t^{3}-9t - 4 = 0\).
• We can use a numerical method (such as the Newton - Raphson method) or a graphing utility to find the roots of the cubic equation \(y = 20t^{3}-9t - 4\).
• Using a graphing calculator or a software like Desmos to find the positive root of the function \(y = 20t^{3}-9t - 4\) (since \(t\gt0\)).
• When we solve the equation \(20t^{3}-9t - 4 = 0\) for \(t\gt0\), we find that \(t\approx0.831\).

Answer:

B. 0.831