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

Question

A) Find the values of t for which the particle is moving to the left.

B) Find the values of t for which the particle is to the left if the origin . Justify your answers for both. Thanks.

Answer 1

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

 

1. 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.  

 

2. 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