Answer to Question #165370 in Physics for Jam Lee

Question

A particle moves along xy-plane with a position of

x (t) = sin(2t) + 3 cos(4t)

y (t) = sin(t) – 6 cos(3t)

a. Determine the position vector of the particle.

b. Determine the velocity vector of the particle.

c. Determine the acceleration vector of the particle.

d. Determine the position vector of the particle when t = 4s.

e. Determine the average acceleration from t = 0s to t= 5s.

f. Determine the instantaneous acceleration when t=Is.

g. Determine the average velocity from t = ls to t= 3s.​

Answer 1

(a) We can write the position vector of the particle moving along xy-plane as follows:

$$r(t)=(sin(2t)+3cos(4t))\hat{i}+(sin(t)-6cos(3t))\hat{j}.$$

(b) To find velocity we need to take the derivative of position function with respect to $t$:

$$v_x(t)=\dfrac{dx(t)}{dt},$$ $$v_x(t)=\dfrac{d}{dt}(sin(2t)+3cos(4t))=2cos(2t)-12sin(4t),$$ $$v_y(t)=\dfrac{dy(t)}{dt},$$ $$v_y(t)=\dfrac{d}{dt}(sin(t)-6cos(3t))=cos(t)+18sin(3t).$$

Finally, we can write the velocity vector of the particle:

$$v(t)=(2cos(2t)-12sin(4t))\hat{i}+(cos(t)+18sin(3t))\hat{j}.$$

(c) To find acceleration we need to take the derivative of velocity function with respect to $t$:

$$a_x(t)=\dfrac{dv_x(t)}{dt},$$ $$a_x(t)=\dfrac{d}{dt}(2cos(2t)-12sin(4t))=-4sin(2t)-48cos(4t),$$ $$a_y(t)=\dfrac{dv_y(t)}{dt},$$ $$a_y(t)=\dfrac{d}{dt}(cos(t)+18sin(3t))=54cos(3t)-sin(t).$$

Finally, we can write the acceleration vector of the particle:

$$a(t)=(-4sin(2t)-48cos(4t))\hat{i}+(54cos(3t)-sin(t))\hat{j}.$$

(d) Let's determine the position vector of the particle when t = 4 s:

$$r(t=4\ s)=(sin(2\cdot4)+3cos(4\cdot4))\hat{i}+(sin(4)-6cos(3\cdot4))\hat{j},$$ $$r(t=4\ s)=-1.88\hat{i}-5.82\hat{j}.$$

(e) Let's first find the velocity vector of the particle at t = 0 s:

$$v(t=0\ s)=(2cos(2\cdot0)-12sin(4\cdot0))\hat{i}+(cos(0)+18sin(3\cdot0))\hat{j}.$$ $$v(t=0\ s)=2\hat{i}+1\hat{j}.$$

Then, we can find the magnitude of the velocity of the particle at t = 0 s from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2)^2+(1)^2}=2.24\ \dfrac{m}{s}.$$

Then, let's find the velocity vector of the particle at t = 5 s:

$$v(t=5\ s)=-12.63\hat{i}+11.98\hat{j}.$$

Then, we can find the magnitude of the velocity of the particle at t = 5 s from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(-12.63)^2+(11.98)^2}=17.41\ \dfrac{m}{s}.$$

Finally, we can find the average acceleration from t = 0 s to t = 5 s:

$$a_{avg}=\dfrac{\Delta v}{\Delta t},$$ $$a_{avg}=\dfrac{17.41\ \dfrac{m}{s}-2.24\ \dfrac{m}{s}}{5\ s}=3.03\ \dfrac{m}{s^2}.$$

(f) Let's find the the instantaneous acceleration when t=1 s:

$$a(t=1\ s)=(-4sin(2\cdot1)-48cos(4\cdot1))\hat{i}+(54cos(3\cdot1)-sin(1))\hat{j},$$ $$a(t=1\ s)=27.74\hat{i}-54.3\hat{j},$$ $$a(t=1\ s)=\sqrt{a_x^2+a_y^2}=\sqrt{(27.74)^2+(-54.3)^2}=61\ \dfrac{m}{s^2}.$$

(g) Let's find the average velocity from t = 1 s to t = 3 s:

$$r(t=1\ s)=(sin(2\cdot1)+3cos(4\cdot1))\hat{i}+(sin(1)-6cos(3\cdot1))\hat{j},$$ $$r(t=1\ s)=-1.05\hat{i}+6.78\hat{j},$$ $$r=\sqrt{r_x^2+r_y^2}=\sqrt{(-1.05)^2+(6.78)^2}=6.86\ m,$$ $$r(t=3\ s)=2.25\hat{i}+5.61\hat{j},$$ $$r=\sqrt{r_x^2+r_y^2}=\sqrt{(2.25)^2+(5.61)^2}=6.04\ m.$$

Finally, we can find the average velocity from t = 1 s to t = 3 s:

$$v_{avg}=\dfrac{\Delta r}{\Delta t},$$ $$v_{avg}=\dfrac{6.04\ m-6.86\ m}{2\ s}=-0.41\ \dfrac{m}{s}.$$