Question

The position of a squirrel running in a park is given by 𝑟 = [(310.280𝑚/𝑠)𝑡 + (0.0360𝑚/𝑠 2 )𝑡 2 ]𝑖̂+ (0.0190𝑚/𝑠 3 )𝑡 3 𝑗̂. (a) What are 𝑣𝑥(𝑡) and 𝑣𝑦(𝑡), the 𝑥- and 𝑦-components of the velocity of the squirrel, as functions of time? (b) At 𝑡 = 5.00 s, how far is the squirrel from its initial position? (c) At 𝑡 = 5.00 s, what are the magnitude and direction of the squirrel’s velocity?

Answer 1

(a)

$$v_x(t)=\dfrac{dr_x}{dt} = (0.28\ \dfrac{m}{s})+(0.072\ \dfrac{m}{s^2})t,$$ $$v_y(t)=\dfrac{dr_y}{dt} =(0.057\ \dfrac{m}{s^3})t^2.$$

(b) Let's first find $x$- and $y$-components of the final position of the squirrel at $t=5.0\ s$:

$$r_x=(0.28\ \dfrac{m}{s})\cdot5.0\ s+0.036\ \dfrac{m}{s^2}\cdot(5.0\ s)^2=2.3\ m,$$ $$r_y=(0.019\ \dfrac{m}{s^3})\cdot(5.0\ s)^3=2.4\ m.$$

Finally, we can find the final position of the squirrel at $t=5.0\ s$ from the Pythagorean theorem:

$$r=\sqrt{r_x^2+r_y^2}=\sqrt{(2.3\ m)^2+(2.4\ m)^2}=3.3\ m.$$

(c) Let's first find $x$- and $y$-components of the squirrel's velocity at $t=5.0\ s$:

$$v_x(t=5.0\ s)= (0.28\ \dfrac{m}{s})+(0.072\ \dfrac{m}{s^2})\cdot5.0\ s=0.64\ \dfrac{m}{s},$$ $$v_y(t=5.0\ s)=(0.057\ \dfrac{m}{s^3})\cdot(5.0\ s)^2=1.43\ \dfrac{m}{s}.$$

We can find the magnitude of the squirrel's velocity from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(0.64\ \dfrac{m}{s})^2+(1.43\ \dfrac{m}{s})^2}=1.6\ \dfrac{m}{s}.$$

We can find the direction of the squirrel's velocity from the geometry:

$$\theta=tan^{-1}(\dfrac{v_y}{v_x}),$$ $$\theta=tan^{-1}(\dfrac{1.43\ \dfrac{m}{s}}{0.64\ \dfrac{m}{s}})=66^{\circ}.$$