Question

What is the average velocity in the time interval t = 2.25 s to t = 6.00 s?

Answer 1

Clearly,the above position-time curve depicts a parabolic relationship,let the equation of parabola be, `x=a(t-h)^2+k` (where, `(h,k)` is the coordinate of vertex)

From the given information, coordinate of vertex is `(8*0.75 , 3*1)` i.e `(6,3)`

And it passes through point, `(2*0.75 , 11*1)` i.e `(1.5 , 11)` (see the beginning point on the left most side of the curve)

So,we can write, `1.5 =a(11-6)^2 + 3`

or, `a=-0.06`

So,the relation between position and time is established as `x=-0.06(t-6)^2 +3`

So,displacement in time `t=2.25` `x=2.15m`
and, at `t=6` `x=3`

So,total displacement in this time interval =`(3-2.15)=0.85m`

So,average velocity = `(0.85)/(6-2.25) m/s =0.23 m/s`

Now, we can rearrange the equation as `x=-0.06t^2+0.72t+0.84`

So,velocity = `(dx)/(dt) = -0.12t +0.72`

So, at `t=3` instantaneous velocity = `-0.12*3 +0.72=0.36 m/s`

Now,let at time `t` velocity `(dx)/(dt)` will be zero,

So, `0=-0.12t +0.72`,

or, `t=0.72/0.12s=6s`

Answer 2

a. -1.65 m/s, b. -2.48 m/s, c. t = 6.00 s

Explanation:

a. We can read the locations at times 2.25 s and 6.00 s from the graph. It is approximate, but I call the locations as follows:

At 2.25 s, the particle was at 8.2 m, at 6.00 s, it was at 2 m.
Therefore the displacement in the specified interval was -6.2 m. The length of the time interval was 3.75 s.

`"Average velocity" = ("total displacement")/("total time")`

`"Average velocity" = (-6.2 m)/(3.75 s) = -1.65 m/s`

b. The slope of the tangent line is calculated `"rise"/"run"`. The rise is actually a fall of 13 m (or a rise of -13 m) and the run is 5.25 s. Therefore, the slope, and the instantaneous velocity at t = 3.00 s, is
`"rise"/"run" = (-13 m)/(5.25 s) = -2.48 m/s`

c. To find the point at which velocity is zero, imagine sliding that tangential line along the curve until the tangential line is horizontal. In that condition, the "rise" would be zero. That happens when t = 6.00 s.

I hope this helps,
Steve