Answer to Question #249267 in Mechanics | Relativity for death

Question

Q.4. 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) $x(2)=2-0.25\cdot2^2=1\ (m)$

$y(2)=1\cdot 2+0.025\cdot2^3=2.2\ (m)$

$s=\int_0^2((0.5t)^2+(1+0.075t^2)^2)^{0.5}dt=2.46\ (m)$

b) $d=\sqrt{1^2+2.2^2}=2.42 (m)$

$v_{avr}=2.42/2=1.21\ (m/s)$

c) $\vec{v}=-0.5t\cdot\vec{i}+(1+0.075t^2)\cdot\vec{j}$

$v_x=-0.5t$ and $v_y=1+0.075t^2$

$|\vec{v}|=\sqrt{(-0.5t)^2+(1+0.075t^2)^2}$