Q50E - University Physics with Modern Physics

Question

A small object moves along the x-axis with acceleration a_x(t) = -(0.0320 m/s³)(15.0 s – t). At t = 0 the object is at x = -14.0 m and has velocity v_0x = 8.00 m/s. What is the x-coordinate of the object when t = 10.0 s?

Step-by-Step Solution

Answer

X-coordinate of the object when t = 10.0 s is 47.33 m.

Step 1: Identification of given values

Acceleration$a_x\left(t\right)=\left(0.0320\mathrm{m}/{\mathrm{s}}^3\right)\left(15.0\mathrm{s}-t\right)$ 

Velocity$V_{\mathrm{ax}}=8.00\mathrm{m}/\mathrm{s}$ 

X-coordinate = -14.0 m (when t = 0).

Step 2: calculation X-coordinate

As we know that,

$$\begin{gathered} a_x\left(t\right)=-\left(0.0320\mathrm{m}/{\mathrm{s}}^3\right)\left(15.0s-t\right) \\ \frac{dv}{dt}=-\left(0.0320\mathrm{m}/{\mathrm{s}}^3\right)\left(15.0s-t\right) \\ dv=-\left(0.0320\mathrm{m}/{\mathrm{s}}^3\right)\left(15.0s-t\right)dt \end{gathered}$$

Now integrating above equation,

$$\begin{gathered} v-v_0=-\left(0.0320\mathrm{m}/{\mathrm{s}}^3\right)\left(15t-\frac{t^2}{2}\right) \\ \frac{dx}{dt}=v_0-\left(0.0320\mathrm{m}/{\mathrm{s}}^3\right)\left(15t-\frac{t^2}{2}\right) \\ dx=v_0=\left(0.0320\mathrm{m}/{\mathrm{s}}^3\right)\left(15t-\frac{t^2}{2}\right)dt \end{gathered}$$ 

Again integrating above equation,

And putting the value of t = 10 s

$$\begin{gathered} x-x_0=v_0{\left(t\right)}_0^{10}-\left(0.0320\mathrm{m}/{\mathrm{s}}^3\right)\left(15{\left(\frac{t^2}{2}\right)}_0^{10}-{\left(\frac{t^3}{6}\right)}_0^{10}\right) \\ x=47.33 \mathrm{m} \end{gathered}$$