Question

Item 4 Constants Part A A small object moves along the z-axis with acceleration ar (t)(0.0320 m/s3) (15.0 s - t). At t 0 the object is at -14.0 m and has velocity ox -8.50 m/s. What is the x-coordinate of the object when t 10.0 s? Express your answer with the appropriate units. LA |z= 1 Value Units Submit Request Answer

Answer 1

We have $a_{x}(t)=-0.032(15-t)$ .

Thus $v_{x}(t)=\int a_{x}(t)dt=0.032t^{2}/2-0.48t+c_{1}$ where c₁ is the constant of integration.

Since at t=0, v_x(t)=8.5 m/s, substituting in the above expression we obtain c₁=8.5 m/s.

Therefore $v_{x}(t)=0.016t^{2}-0.48t+8.5$ .

Integrating once more we obtain

$x(t)=\int v_{x}(t)dt=0.16t^{3}/3-0.48t^{2}/2+8.5t+c_{2}$ where c₂ is the second constant of integration.'

We are given x(0)=-14.0 m. Substituting in the above expression we obtain:

c₂=-14.0 m.

Therefore we have the final expression

$x(t)=0.016t^{3}/3-0.48t^{2}/2+8.5t-14$.

At t=10 sec, $x(t)=0.016*10^{3}/3-0.48*10^{2}/2+8.5*10-14=52.3 m.$