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^.

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

A) What is , the y-component of the velocity of the squirrel, as function of time?

B) At 5.88 s , how far is the squirrel from its initial position?

C) At 5.88 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

a) vy(t)=(0.0570m/s3)t2 b) vy(t)=(0.0570m/s3)t+(0.0720m/s2)t2 c) vy(t)=(0.0570m/s2)t2 d) vy(t)=(0.0570m/s3)t

Explanation / Answer

V = dR/dt

V =  = d[(0.280m/s)t+(0.0360m/s2)t^2]i+ (0.0190m/s3)t^3j ]/dt

V = (0.280+0.0720t )i+ 0.0570t^2 j

Vx = (0.280+0.0720t )

Vy = 0.0570t^2

B) at t = 5.88 sec

R = (0.280*5.88 +0.0360*(5.88)^2 )i+ 0.0190*(5.88)^3

magnitude of R = sqrt (0.280*5.88+0.0360*(5.88)^2 )^2+ (0.0190*(5.88)^3)^2  = 4.82 m

C)

V = (0.280+0.0720*5.88)i+ 0.0570(5.88)^2 j

angle is 70.41 counterclock wise from x axis

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