Answer to Question #169140 in Physics for mahikah

Question #169140

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 turtleโ€™s position as a function of time is ๐‘ฅ(๐‘ก) = 50.0 ๐‘๐‘š +

(2.00 ๐‘๐‘š/s) โˆ’ t (0.0625 ๐‘๐‘š/ s2) t2

.(a) Find the turtleโ€™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?


1
Expert's answer
2021-03-08T08:27:40-0500

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


"x'(t)=v(t)=2\\text{ cm\/s}-0.0625\\text{ cm\/s}^2 \u00b7t"

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\\\\\nt=\\frac{2}{0.0625}=32\\text{ s}."

(c)The starting point is 50 cm:


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

(d) Substitute this into x(t):


"10=50+2t-0.0625t^2,\\\\\n0.0625t^2-2t-40=0,\\\\\nt=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\u00b745.9=-0.0869\\text{ m\/s}."


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