Answer to Question #169140 in Physics for mahikah

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) = 50.0 𝑐𝑚 +

(2.00 𝑐𝑚/s) − t (0.0625 𝑐𝑚/ s²) t²

.(a) Find the turtle’s initial velocity, initial position, and initiaal acceleration

(b) At what time t is the velocity of the turtle zero?

(c) How long after startingd doesit take the turtle to return to its starting point?

(d) At what times t is the turtle a distance of

10.0 cm from its starting point?

(e) What is the velocity (magnitude and direction) of the turtlea at each of these times?

Answer 1

(a) Find the first derivative of x(t):

$$x'(t)=v(t)=2\text{ cm/s}-0.0625\text{ cm/s}^2 ·t$$

The initial velocity at t=0 is 2 cm/s, initial acceleration is 0.0625 cm/s², initial position at t=0 is 50 cm.

(b) Find when the velocity is 0:

$$v(t)=0=2-0.0625t,\\\space\\ t=\frac{2}{0.0625}=32\text{ s}.$$

(c)The starting point is 50 cm:

$$50=50+2t-0.0625t^2,\\ t=32\text{ s}.$$

(d) Substitute this into x(t):

$$10=50+2t-0.0625t^2,\\ 0.0625t^2-2t-40=0,\\ t=45.9\text{ s}.$$

(We take only positive root because we can't turn back time)

(e) At t=0 it is 2 cm/s, at t=32 is is 0, at t=45.9 it is

$$v(45.9)=2-0.0625·45.9=-0.0869\text{ m/s}.$$