Answer to Question #249267 in Mechanics | Relativity for death

Question #249267

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.


1
Expert's answer
2021-10-10T16:06:09-0400

a) x(2)=20.2522=1 (m)x(2)=2-0.25\cdot2^2=1\ (m)

y(2)=12+0.02523=2.2 (m)y(2)=1\cdot 2+0.025\cdot2^3=2.2\ (m)


s=02((0.5t)2+(1+0.075t2)2)0.5dt=2.46 (m)s=\int_0^2((0.5t)^2+(1+0.075t^2)^2)^{0.5}dt=2.46\ (m)


b) d=12+2.22=2.42(m)d=\sqrt{1^2+2.2^2}=2.42 (m)


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


c) v=0.5ti+(1+0.075t2)j\vec{v}=-0.5t\cdot\vec{i}+(1+0.075t^2)\cdot\vec{j}


vx=0.5tv_x=-0.5t and vy=1+0.075t2v_y=1+0.075t^2


v=(0.5t)2+(1+0.075t2)2|\vec{v}|=\sqrt{(-0.5t)^2+(1+0.075t^2)^2}


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