Question

A amall object moves along the x axsis whit acceleration ax(t)=-0.0320m/s³(15s-t) at t=0 the object is at the x=-14m and has velocity v0z=8m/s. what is the x cordinate of the object when t=10s?

Answer 1

To find the x-coordinate of the object at t=10s, we first integrate the given acceleration function to get the velocity and then integrate again to find the position. Plugging in the initial conditions provides the position equation, which we use to calculate the position at t=10s.

Step-by-step explanation:

To determine the x-coordinate of the object at t=10s, we will integrate the acceleration function to find the velocity function and then integrate again to find the position function. Given that the acceleration is ax(t) = -0.0320 m/s³ (15s - t), we integrate this to find the velocity.

The velocity function v(t) becomes:

$v(t) = -0.0320 m/s³ (15s \cdot t - \frac{t^2}{2}) + C$

We solve for C using the initial velocity v0x = 8 m/s at t=0.

$8 m/s = -0.0320 \cdot 0 + C \Rightarrow C = 8 m/s$

Therefore, the velocity function is:

$v(t) = -0.0320 m/s³ (15s \cdot t - \frac{t^2}{2}) + 8 m/s$

Next, we integrate the velocity function to find the position function x(t).

The position function x(t) is:

$x(t) = \int (-0.0320 m/s³ (15s \cdot t - \frac{t^2}{2}) + 8 m/s) dt$

Integrating, we get:

$x(t) = -0.0320 m/s³ (\frac{15s \cdot t^2}{2} - \frac{t^3}{6}) + 8t + D$

We find D using the initial position x0 = -14 m at t=0.

$x(0) = -14 m = 0 + D \Rightarrow D = -14 m$

The position function is:

$x(t) = -0.0320 m/s³ (\frac{15s \cdot t^2}{2} - \frac{t^3}{6}) + 8t - 14 m$

Finally, to find the x-coordinate at t=10s, we substitute t with 10s in the position function:

$x(10s) = -0.0320 m/s³ (\frac{15 \cdot 100}{2} - \frac{1000}{6}) + 8 \cdot 10 - 14 m$

After solving this equation, we obtain the x-coordinate of the object at t=10s.