Question

A particle starts at time t=0 and moves along a number line so that its position, at time t is greater than or equal to zero, is given by x(t)= (t-2)(t-6)^3. The particle is moving to the left for...

Answer 1

What a strange cookie. We need to study the critical points for this one.

As we can see, at t=0 (at the start) : x(0) = (-2)* (-6)^3= 432. So that's our initial position.

At t=2, x(2)=0. We can see that our particle is moving towards the left on the axis (by getting from 432 to 0)

At t>2 and t<6, x(t) <0. Our particle is still going left - now beyond the y axis.

At t=6, x(6) = 0 again but since we were left from the y axis (because we were going negative earlier) that means the particle took a right turn this time.

At t>6, x(t)>0 the particle is still going right and will keep on going to infinity.

So the answer will be : 6 seconds.