The position of a squirrel running in a park is given by π = [(310.280π/π )π‘ + (0.0360π/π 2 )π‘ 2 ]πΜ+ (0.0190π/π 3 )π‘ 3 πΜ. (a) What are π£π₯(π‘) and π£π¦(π‘), the π₯- and π¦-components of the velocity of the squirrel, as functions of time? (b) At π‘ = 5.00 s, how far is the squirrel from its initial position? (c) At π‘ = 5.00 s, what are the magnitude and direction of the squirrelβs velocity?
(a)
"v_x(t)=\\dfrac{dr_x}{dt} = (0.28\\ \\dfrac{m}{s})+(0.072\\ \\dfrac{m}{s^2})t,""v_y(t)=\\dfrac{dr_y}{dt} =(0.057\\ \\dfrac{m}{s^3})t^2."(b) Let's first find "x"- and "y"-components of the final position of the squirrel at "t=5.0\\ s":
Finally, we can find the final position of the squirrel at "t=5.0\\ s" from the Pythagorean theorem:
(c) Let's first find "x"- and "y"-components of the squirrel's velocity at "t=5.0\\ s":
We can find the magnitude of the squirrel's velocity from the Pythagorean theorem:
We can find the direction of the squirrel's velocity from the geometry:
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