Question #95482
A snowmobile is originally at the point with position vector 28.5 m at 95.0° counterclockwise from the x axis, moving with velocity 4.69 m/s at 40.0°. It moves with constant acceleration 1.85 m/s2 at 200°. After 5.00 s have elapsed, find the following. (Express your answers in vector form.)
1
Expert's answer
2019-09-30T10:06:29-0400

After 5 seconds velocity is given by

v=u + 5a\overset{\to}{v}=\overset{\to}{u}\ +\ 5\overset{\to}{a}

Talking about components

vx=ux + 5axvy=uy + 5ayv_x=u_x\ +\ 5a_x\\v_y=u_y\ +\ 5a_y

vx=4.69cos(40°)+5×1.85cos(200°)vy=4.69sin(40°)+5×1.85sin(200°)v_x=4.69\cos(40\degree)+5\times1.85\cos(200\degree)\\v_y=4.69\sin(40\degree)+5\times 1.85\sin(200\degree)

vx=5.098vy=0.149v=vx2+vy2=5.1 m sec1v_x=-5.098\\v_y=-0.149\\|v|=\sqrt{v_x^2+v_y^2}=5.1\ m\ sec^{-1}

tan(θ)=0.1495.098=0.0292θ=1.67since both component are negative soangle will be 180+1.67=181.67°\tan(\theta)=\frac{0.149}{5.098}=0.0292\\\theta=1.67\\since \ both\ component\ are\ negative\ so angle \ will\ be\ 180+1.67=181.67\degree So velocity after 5 seconds is elapsed will be

5.1 m sec15.1\ m\ sec^{-1} at an angle of 181.67°181.67\degree

Or in term of vector component

v=5.098i^0.149j^\overset{\to}{v}=-5.098\hat{i}-0.149\hat{j}


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