Answer to Question #249166 in Mechanics | Relativity for Ruby

Question

A robot lands on Mars. Its x and y coordinates are given by equations: x(t) = 2m –(0.25m/s2 )t2

y(t)= (1.0 m/s)t +(0.025m/s3)

(a).What is the position and distance from landing at t=2 seconds?(b)What is the displacement and average velocity during the interval from t=0 s to t= 2?(c).Derive a general expression for its instantaneous velocity vector and find its instantaneous velocity in its component form as well as in terms of its magnitude.

Answer 1

(a) The position and distance are

$$x(t) = 2 –(0.25)t^2,\\ x(2)=1\text{ m},\\ y(t)= (1.0)t +(0.025)t^3,\\ y(2)=2.2\text{ m},\\ d=\sqrt{y^2+x^2}=2.42\text{ m}.$$

(b) The displacement is d, or 2.42 m, the average velocity is

$$v=\frac dt=1.21\text{ m/s}.$$

(c) In component form:

$$v_x(t)=x'(t)=-0.5t,\\ v_y(t)=y'(t)=1+0.075t^2.\\ |v|=\sqrt{v_x^2(t)+v_y^2(t)}=\\ =\sqrt{1+0.4t^2+5.625·10^{-3}t^4}.$$