a particle moves along the x - axis so that its position at time t > 0 is given by x(t) and…
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 = \na 0.387\nb 0.831\nc 1.243\nd 1.647\ne - 0.094
Answer
- 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).
- 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).