Asmall object moves along the xx-axis with acceleration ax(t)ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3) (15.0s−t). at tt = 0 the object is at xx = -14.0 mm and has velocity v0xv0x = 8.70 m/sm/s.

what is the xx-coordinate of the object when tt = 10.0 ss?

x = 54.3m (on the +ve x axis)

Explanation:

This is an Initial Value Problem. That means the initial values of certain parameters have been given and that can help solve the problem.

Given that acceleration, a, is:

ax(t) = - 0.032(15.0 - t)

And the initial values are:

x(t = 0) = - 14.0m

v(t = 0) = 8.7m/s

Hence,

a = - 0.032(15 - t)

a = - 0.48 + 0.032t

a = dv/dt = -0.48 + 0.032t

To obtain the velocity, v, integrate the acceleration and apply the initial values of v and t:

v = ∫dv/dt  = ∫(-0.48 + 0.032t)

∫dv = ∫(-0.48 + 0.032t)dt

(v - v₀) = -0.48(t - t₀) + 0.032(t²/2 - t₀²/2)

Inputting the initial values t₀ = 0s, v₀ = 8.7m/s:

=> v - 8.7 = -0.48t + 0.032t²/2

v = 8.7 - 0.48t + 0.016t²

To obtain distance, x, integrate the velocity and apply the initial values:

v = dx/dt = 8.7 - 0.48t + 0.016t²

=> ∫dx/dt = ∫(8.7 - 0.48t + 0.016t²)

∫dx= ∫(8.7 - 0.48t + 0.016t²)dt

(x - x₀) = 8.7(t - t₀) - 0.48(t²/2 - t₀²/2) + 0.016(t³/3 - t₀³/3)

Inputting the initial values t₀ = 0s, x₀ = - 14.0m:

(x + 14.0) = 8.7t - 0.48t²/2 + 0.016t³/3

x = 8.7t - 0.48t²/2 + 0.016t³/3 - 14.0

Now that the distance, x, has been obtained, when t = 10s:

x = 8.7*10 - 0.48*10²/2 + 0.016*10³/3 - 14.0

x = 87 - 24 + 5.3 - 14.0

x = 54.3m

Therefore, at time, t = 10s, x = +54.3m. (i.e. 54.3 on the +ve x axis).

part 1: which substance has a higher solubility?

a

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