Particle Motion (Again) [Re-typed]

Question

A particle moves along a straight line with velocity given by v(t) = 5-(1.03)^-t^2 at time t ≥ 0. What is the acceleration of the particle at time t = 6?

Keep in mind: (1.03) is to the -t power and that -t is to the 2nd power.

Answer 1

v(t) = 5-1.03^{-t^2}

a(t) = dv/dt

Need to use implicit derivation with logs

v-5 = 1.03^{-t^2}

ln(v-5) = -t²ln(1.03)

Take derivative of both sides

(1/(v-5))dv/dt = -2tln(1.03)

dv/dt = a(t) = -2tln(1.03)(v-5) we know v = 5-1.03^{-t^2}

= -2tln(1.03)(5-1.03^{-t^2}-5)

= 2tln(1.03)(1.03^{-t^2})

a(6) = 12ln(1.03)(1.03^-36) = 12(0.03)(0.345) = 0.1242

Answer 2