Answer to Question #259168 in Physics for OMAR

Question

The position of a particle moving along the x-axis is given by 

x = 3.0t² - 2.0t³ where x is in meters and t is in seconds. What is the position 

of the particle when it achieves its maximum speed in the positive x-direction?

I think the answer is 1 m but in the solutions manual there is 0.50 m.

Please tell me why 1 m is wrong. If we use a graphing calculator we will see that maximum point at 1 s and its value is 1 m.

Answer 1

The velocity of the particle is given as follows:

$$v(t) = x'(t) = 6t - 6t^2$$

The maximum value of velocity is achieved when its derivative is equal to 0:

$$v'(t) = 6-12t_{max}=0\\ t_{max} = 0.5s$$

The graph of $v(t)$ confirms that indeed, velocity has maximum at 0.5s.

Substituting $t_{max}$ to $x(t)$, get:

$$x(t_{max}) = 3\cdot 0.5^2 - 2\cdot 0.5^3 = 0.5m$$

Answer. 0.5m