Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.40 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer 1

The x-coordinate of the object when t = 10.0 s is -13.6 m.

Step-by-step explanation:

To find the x-coordinate of the object at t = 10.0 s, we need to find the position function x(t) first. To do this, we integrate the acceleration function ax(t) with respect to t twice. First, we integrate ax(t) to get the velocity function vx(t), which will give us the x-component of the velocity at any given time:

vx(t) = ∫ ax(t) dt = ∫ -(0.0320m/s^3)(15.0s-t) dt = -0.0320(t - 7.5)^2 + C

Next, we integrate vx(t) to get the position function x(t), which will give us the x-coordinate of the object at any given time:

x(t) = ∫ vx(t) dt = ∫ (-0.0320(t - 7.5)^2 + C) dt = -0.0320(t - 7.5)^3/3 + C(t - 7.5) + D

Now, we can plug in the given values to find the specific x-coordinate at t = 10.0 s. Since the object is at x = -14.0 m when t = 0 and has a velocity of v0x = 7.40 m/s, we can set up two equations:

-14.0 = -0.0320(0 - 7.5)^3/3 + C(0 - 7.5) + D

7.40 = -0.0320(0 - 7.5)^2 + C

Solving for C and D, we find C = 0.427 and D = -14.0. Plugging these values into the position function, we get:

x(t) = -0.0320(t - 7.5)^3/3 + 0.427(t - 7.5) - 14.0

Finally, plugging in t = 10.0 s, we can find the x-coordinate of the object:

x(10.0) = -0.0320(10.0 - 7.5)^3/3 + 0.427(10.0 - 7.5) - 14.0 = -13.6 m