Aparticle'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 angle of the particle's displacement vector δr⃗ relative to the point r⃗ 0=(0.0i^+7.0j^)m.
express your answer using two significant figures.

The given particle position is


When t=5 s, the position vector is


The reference vector at t = 5 s is


Let θ = the angle between the two vectors.
Then, by definition,


|r₁} =  √[175²+(-368)²] = 407.4911
|r₂| = 7


Therefore
θ = cos⁻¹ -2576/(407.4911*7) = cos⁻¹ -0.9031 = 154.57°

154.57°