let s(t)=t^2-t-2 be an equation for the position of a particle moving along a horizontal path

Question

A )The values of t for which the particle is moving to the left . Justify your answer. B) The values of t for which the particle is to the left of the origin (position zero) Justify your answer.

Answer 1

A) s'(t)<0 when particle moves to left: 2t-1<0, t<1/2. Rate of change of s (ds/dt or s') is velocity and negative velocity is movement to left.

B) s(t)<0 when particle is positioned left of zero: t^2-t-2<0; (t-2)(t+1)<0. When t<2, s(t)<0. (Assume t can't be negative.)

Answer 2

The position of the particle is given by s(t) = t^2 - t - 2

A) The velocity of the particle, v(t) is given by the first derivative of the position: v(t) = 2t -1
When the velocity is >0, the particle is moving to the right. When the velocity is <0 the particle is moving to the left.
So, you want 2t-1 < 0, 2t < 1, t< 1/2

I'm assuming that you start monitoring the position at t=0, so the time when the particle is moving to the left is 0<t<1/2

If you don't want to use calculus, you can do it this way:
Factor the position: s(t) = (t-2)(t+1). This is an upward parabola with zeroes at t=2 and t-1. The vertex of the parabola is halfway between these two: (2+-1) /2. = 1/2. This is when the parabola is at a minimum, when t is less than this value, s(t) is decreasing, i.e. moving to the left. When t is greater than this value, s(t) is increasing, i.e. moving to the right.

B) When the particle is to the left of the origin, s(t) is <0.
since s(t) = (t-2) (t+1), s(t) is less than zero when either t-2 < 0, or t+1 <0, but NOT both.

When t-2 < 0, this means t<2, and the first term is <0. However, once t<-1, then both terms are <0, and their product is >0. So, the only time that only one of them is <0 is in between these two values, i.e.

-1 < t <2. Since we only care about times >0, s(t) will be to the left of the origin for 0 < t <2

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