Raul started out on his bike thirty meters to the south of my position. Five seconds later, he was twenty meters to the north. How fast is Raul biking. Write
an equation to find his displacement, x, from the time, t. If he keeps going at the
same rate, where will he be one minute (sixty seconds from) when he started?

Let's define the North as the positive y-axis.

At t = 0s (t is the variable of time) we know that Raul is 30m at the South, so in our system, the position of Raul at t = 0s is:

p(0s) = -30m

After 5 seconds, at the time t = 5s, we know that the position of Raul is 20m at the North, so now the position is:

p(5s) = 20m.

a) First, speed is calculated as distance over time, or:

S = ( p(5s) - p(0s))/5s = (20m - (-30m))/5s = 50m/5s = 10m/s.

b) The displacement is defined as the difference between the final position and the initial position:

Displacement = p(5s) - p(0s) = 20m - (-30m) = 50m.

Now, knowing the speed is 10m/s, we can write the movement equation as:

p(t) = 10m/s*t + p(0s) = 10m/s*t - 30m

So the displacement at a time t is:

D = p(t) - p0s = 10m/s*t - 30m - (-30m) = 10m/s*t

c) the position one minute after he started will be:

p(1min)

But we have our equation in seconds, so we write 1min = 60s

p(60s) = 10m/s*60s - 30m = 600m - 30m = 570m

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by