Chapter 2.4, Problem 30E

To determine

The time at which the particle is at the point $\left(5,2\right)$ .

Answer to Problem 30E

It has been determined that the particle is at the point $\left(5,2\right)$ after $2.832$ units of time from the start of its motion.

Explanation of Solution

Given:

The parametric equations of motion for a particle are $x\left(t\right)=4t^3-16t^2+15t$ and $y\left(t\right)=2$ for $t\ge 0$ .

Concept used:

The time at which the particle is at the given point can be obtained by equating the parametric equations to the respective coordinates of the point and then solving for $t$ .

Calculation:

Let the particle be at the point $\left(5,2\right)$ at time $t$ .

Then, $x\left(t\right)=5$ and $y\left(t\right)=2$ .

Using the expressions for the parametric equations, it follows that:

  $4t^3-16t^2+15t=5$ and $2=2$

The second equation is an identity and does not give any value of $t$ .

Solving the first equation,

  $4t^3-16t^2+15t=5$

Simplifying,

  $4t^3-16t^2+15t-5=0$

Now, the only real root of this equation is $t=2.832$ .

This implies that the particle is at the point $\left(5,2\right)$ after $2.832$ units of time from the start of its motion.

Conclusion:

It has been determined that the particle is at the point $\left(5,2\right)$ after $2.832$ units of time from the start of its motion.