Rabbit run comprehension:
A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn.
The coordinates (meters) of the rabbit's position as functions of time t (seconds) are given by

$x=-{t^2\over 2}+5t+20$
And $y=t^2-10t+30$

Find the acceleration at time t = 15 s?

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Solution

The correct option is D

So the particle position vector in x direction is give by
$x (t) = \frac{- t^{2}}{2} + 5 t + 20 v_{x} (t) = \frac{d x}{d t} = - t + 5 a_{x} (t) = \frac{d v}{d t} = 1^i N o w p o s i t i o n v e c t o r i n y d i r e c t i o n i s g i v e n b y y (t) = t^{2} - 10 t + 30 v_{y} (t) = \frac{d y}{d t} = 2 t - 10 a_{y} (t) = \frac{d v_{y}}{d t} = 2^j$

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Similar questions

A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn.
The coordinates (meters) of the rabbit's position as functions of time t (seconds) are given by

$x = - \frac{t^{2}}{2} + 5 t + 20$
And $y = t^{2} - 10 t + 30$

Find the acceleration at time t = 15 s?

A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn.
The coordinates (meters) of the rabbit's position as function of time t(seconds) are given by

$x (t) = - \frac{t^{2}}{2} + 5 t + 20$
And $y (t) = t^{2} - 10 t + 30$

Find the velocity at time t = 15s.

A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn.
The coordinates(meters) of the rabbit's position as function of time t(seconds) are given by

$x (t) = - \frac{t^{2}}{2} + 5 t + 20 A n d y (t) = t^{2} - 10 t + 30$

(a)At t=10s, what is the rabbot's position vector $\to r$ in unit-vector notation and in magnitude-angle notation?

A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn.
The coordinates(meters) of the rabbit's position as function of time t(seconds) are given by

$x (t) = - \frac{t^{2}}{2} + 5 t + 20 A n d y (t) = t^{2} - 10 t + 30$

(a)At t=10s, what is the rabbot's position vector $\to r$ in unit-vector notation and in magnitude-angle notation?

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The correct option is D So the particle position vector in x direction is give by x(t)=−t22+5t+20vx(t)=dxdt=−t+5ax(t)=dvdt=1^iNow position vector in y direction is given byy(t)=t2−10t+30vy(t)=dydt=2t−10ay(t)=dvydt=2^j