Particle Motion (Again)

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?

Answer 1

Hi Hart,

So we will need to find the derivative of v(t) =, because a(t)=v'(t)

v(t)=5-(1.03*t^2)

v'(t)=0+2.06t (using the power rule)

now we plug in 6 for t into v't=a(t)

a(6)=2.06*6=12.36.

So the acceleration at t=6 is 12.36.