A particle's position as a function of time t is given by r⃗ =(5.0t+6.0t2)mi+(7.0−3.0t3)mj.
At t=5.0s, find the magnitude of the particle's displacement vector Δr⃗ relative to the point r⃗ 0=(0.0i+7.0j)m.

At t=5.0s, find the angle of the particle's displacement vector Δr⃗ relative to the point r⃗ 0=(0.0i+7.0j)m.

Help please i'm stuck

Question

A particle's position as a function of time t is given by r⃗ =(5.0t+6.0t2)mi+(7.0−3.0t3)mj.                           
At t=5.0s, find the magnitude of the particle's displacement vector Δr⃗  relative to the point r⃗ 0=(0.0i+7.0j)m.

At t=5.0s, find the angle of the particle's displacement vector Δr⃗  relative to the point r⃗ 0=(0.0i+7.0j)m.

Help please i'm stuck

loser66 · Answer

$\vec r_0$ means the position of the particle at time t =0 . It is  $\vec r_0= (0.0 i + 7.0 j)m$
at t=5.0s, the position of it is $\vec r_5= (5.0*5) i + (6.0*(5^2))j= (25.0 i +150 j)$
                                                             ---------------------------------------
                                                                      $\triangle r =r_5 - r_0 =  25.0i +143 j)$