The position of a squirrel running in a park is given by r =[(0.280m/s) t +(0.03

Question

The position of a squirrel running in a park is given by r =[(0.280m/s)t+(0.0360m/s2)t2]i^+ (0.0190m/s3)t3j^.

A. What is x(t), the x-component of the velocity of the squirrel, as function of time?

B. What is y(t), the y-component of the velocity of the squirrel, as function of time?

C. At 4.30 s , how far is the squirrel from its initial position?

D. At 4.30 s , what is the magnitude of the squirrel's velocity?

E. At 4.30 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

Explanation / Answer

.v = dr/dt = [0.280 + 0.072t]i + 0.057t^2j

Part (A)

the x-component of the velocity of the squirrel, as function of time

= [0.280 + 0.072t]i

Part (B)

the y-component of the velocity of the squirrel, as function of time

= 0.057t^2j

Part (C)

r (4.30)= [(0.280m/s)(4.30)+(0.0360m/s2)(4.30)^2]i^+ (0.0190m/s3)(4.30)^3j

r (4.30)= 1.87i + 1.51j

d= sqrt (i^2+ j^2) = ((1.87)^2 + (1.51)^2)= 2.40 m

Part (D)

v(4.30) = [0.280 + 0.072x 4.30]i + 0.057 x 4.30^2j

= 0.59i + 1.05 j

magnitude = sqrt( (0.59)^2 + (1.05)^2) = 1.20 m/s

Part (E)

direction = tan-1(1.05/0.59) = 60.67 degrees

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