Question

A rabbit runs across a parking lot The coordinates of the rabbit as a function of the time t x =-0.3t^2+7.2t+28 and y = 0.22 +² - 9.1t+ 30 with t in secs and x in metres At t=15s, what is the rabbit's position vector r in unit vector notation and as a magnitude angle​

Answer 1

Rabbit's position vector (r) in the unit vector is 203.5î - 57j and the magnitude and directional angle of vector (r) are 195.35m and 0.004° respectively.

Given

x = 0.3t² + 7.2t + 28

y = 0.22t² - 9.1t +30

At t = 15s

x = 0.3(15)² + 7.2(15) +28 = 203.5

y = 0.22(15)² - 9.1(15) + 30 = -57

The rabbit's position vector (r) in unit vector notation is:

r = 203.5î - 57j

The magnitude of position vector (r)

r = √x² + y²

r = √(203.5)² - (57)² =√38163.25

  =  195.35m

The directional angle of position vector (r)

tanθ = y/x = -57/203.5

θ = $tan^{-1}$ (-57/203.5)

θ = 0.004°

Hence, rabbit's position vector (r) in the unit vector is 203.5î - 57j and the magnitude and directional angle of vector (r) are 195.35m and 0.004° respectively.

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