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A rabbit runs across a parking lot on wh...

A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn . The coordinates (meters) of the rabbits position as functions of time t (seconds) are given by
`x = -0.31t^2 + 7.2 t + 28`
and `y = 0.22 t^2 - 9.1 t + 30`
For the rabbit in the precending sample problem , find the velocity `vec v` at time t = 15s .

Text Solution

Verified by Experts

We can find `vecv` by taking derivatives of the components of the rabbit.s position vector .
Applying the `v_x` part of Eq. We find the x component of `vecv` to be
`v_x = (dx)/(dt) = (d)/(dt) (-0.31t^2 + 7.2 t + 28)`
` = -6.02 t + 7.2 `
At t = 15s , this gives `v_x = -2.1 m//s` similarly , applying the `v_y` part of eq. we find
`v_y = (dy)/(dt) = (d)/(dt) (0.22 t^2 - 9.1 t + 30)`
` = 0.44 - 9.1 `
At t = 15 s , this gives `v_y` = -2.5 m/s eqn. then yields
`vec v = (-2.1 m//s) hati + (-2.5m//s) hatj `
tangent to the rabbits path and in the direction the rabbit is running at t = 15s. to get the magnitude and angle of `vecv` either we use eq. to wire
`v = sqrt(v_x^2 + v_y^2) = sqrt( (-2.1 m//s)^2 + (-2.5 m//s)^2)`
` = 3.3 m//s `
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