Question

A particle moves in the xy plane with constant acceleration. At time t=0 s, the position vector for the particle is r=9.70mx^+4.30my^. The acceleration is given by the vector a=8.00m/s^2x^+3.90m/s^2y^. The velocity vector at time t=o s is v=2.80m/sx^ - 7.00m/sy^.

What is the magnitude of the position vector at time t= 2.10 s?

What is the angle between the position vector and the positive x-axis at time t= 2.10 s?

Answer 1

(i)

$x(t)=x_0+v_{x0}t+\frac {1}{2}a_xt^2$

$x(t)=9.70+2.80\cdot 2.10+\frac {1}{2}\cdot 8\cdot 2.10^2=33.22\: m$

$y(t)=y_0+v_{y0}t+\frac {1}{2}a_yt^2$

$y(t)=4.30+(-7)\cdot 2.10+\frac {1}{2}\cdot 3.90\cdot 2.10^2=-1.80\: m$

$|\vec r(t=2.10\: s)|=\sqrt {33.22^2+(-1.80)^2}\: m= 33.26\: m$

(ii)

$\theta =270+90-\tan ^{-1}\frac {1.80}{33.22}=356.89^{\degree}$