Answer to Question #300525 in Physics for Jerry!!!123???

Question

A turtle crawls along a straight line, which we will call the x-axis with the

positive direction to the right. The equation for the turtle’s position as a function of time is x(t) = 51.0 cm +

(2.50 cms) t − (0.0623 cms2) t2

a.)At what time t is the velocity of the turtle zero?

(b) How long after starting does it take the turtle to return to its starting point?

(c) At what times t is the turtle a distance of 0.1 m from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times?

Answer 1

Given:

$x(t) ={ 51.0\: cm} +(2.50\: cm/s) t − (0.0623\: cm/s^2) t^2$

(a)

$$v=x'(t) =(2.50\: cm/s) − (0.1246\: cm/s^2) t=0$$

$$t=\frac{2.50\: cm/s}{0.1246\: cm/s^2}=20\:\rm s$$

(b)

$${ 51.0\: cm} +(2.50\: cm/s) t − (0.0623\: cm/s^2) t^2=0$$

Root: $t=55\:\rm s$

(c)

$${ 51.0\: cm} +(2.50\: cm/s) t − (0.0623\: cm/s^2) t^2=10$$

Root: $t=52.6\:\rm s$