I have this vector that is confusing me I know what I need to do but the components of these vectors are confusing me.

Problem:
You are standing at point with coor dinates (1m, -1m, -2m) another erson is at r(t)=(8m)i - (3m/s^3) t^3 j + (4m/s^2) t^2 k Find distance between you and other person at t=1s.

How do I deal with those m and m/s^3 in the components

Thanks

find position at t=1 of the other person.

Then, distance=sqrt((x-xo)^2 + (y-y0)^2 + (z-zo)^2 )

To find the distance between two points in space, we can use the formula:

d = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)

In this case, let's consider your coordinates as (x1, y1, z1) = (1m, -1m, -2m).

Now, let's focus on the other person's position vector r(t) = (8m)i - (3m/s^3) t^3 j + (4m/s^2) t^2 k. At t = 1s, we can substitute t = 1 into the equation to find the other person's position:

r(1s) = (8m)i - (3m/s^3)(1s)^3 j + (4m/s^2)(1s)^2 k

To simplify the units, let's break it down step by step:

1. (8m)i - (3m/s^3)(1s)^3 j = (8m)i - (3m/s^3) j
- The (8m)i component remains unchanged.
- For the j component, the unit (3m/s^3) is raised to the power of 3 since it multiplies t^3. This results in (3m/s^3)(1s)^3 = 3m/s^3.

2. (8m)i - (3m/s^3) j + (4m/s^2)(1s)^2 k = (8m)i - (3m/s^3) j + (4m/s^2) k
- The (4m/s^2) component does not need any changes.

Now, we have the coordinates for the other person at t = 1s as (x2, y2, z2) = (8m, -3m/s^3, 4m/s^2).

Substituting these values into the distance formula, we get:

d = √((8m - 1m)^2 + (-3m/s^3 + 1m)^2 + (4m/s^2 + 2m)^2)

Simplifying further, we have:

d = √((7m)^2 + (4m/s^3)^2 + (6m)^2)

d = √(49m^2 + 16m^2/s^6 + 36m^2)

d = √(49m^2 + 16m^2/s^6 + 36m^2) = √(85m^2 + 16m^2/s^6)

Therefore, the distance between you and the other person at t = 1s is approximately √(85m^2 + 16m^2/s^6).