A particle moves in the xy-plane

Question

A particle moves in the xy-plane so that its velocity vector at time t is v(t) = <t², sin(πt> and the particle’s position vector at time t = 0 is <1,0>. What is the position vector of the particle when t = 3?

a) <9, 1/π>

b) <10, 2/π>

c) <10, 2>

d) <10, 2π>

e) <10, -2/π>

Answer 1

Since v(t) = <t², sin(πt>, the velocity in the x-direction is t² and the velocity in the y direction is sin(πt):

V_x = t²

V_y = sin(πt)

V=ds/dt

Vdt = ds

∫ds = ∫Vdt

S = ∫Vdt

S_x = ∫V_xdt

S_x = ∫t²dt = 1/3t³ + C₁

Since at t = 0, the position is <1, 0>, then S_x(0) = 1 so:

S_x(0) = 1 = 1/3(0)³ + C₁ so C₁ = 1 making our complete S_x(t) function:

S_x(t) = 1/3t³ + 1

For t = 3: S_x(3) = 1/3(3)³ + 1 = 10

S_y = ∫V_ydt = ∫sin(πt) dt = -1/π(cos(πt) + C₂

Since at t = 0, the position is <1, 0>, then S_y(0) = 0 so:

S_y = 0 = -1/π(cos(π(0)) + C₂ so C₂ = 1/π making our complete S_y(t) function:

S_y(t) = -1/π(cos(πt) + 1/π

For t = 3: S_y(3) = -1/π(cos(π(3)) + 1/π = 2/π

So at t = 3, the position <S_x, S_y> = <10, 2/π>