Question

After a ball rolls off the edge of a horizontal table at time t = 0, its velocity as a function of time is given by

$\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\hat{\mathbf{i}}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{8}\mathit{t}\hat{\mathbf{j}}$

where$\overrightarrow{\mathbf{v}}$is in meters per second andis in seconds. The ball's displacement away from the edge of the table, during the time interval of 0.380 sfor which the ball is in flight, is given by

$\mathit{\Delta }\overrightarrow{\mathbf{r}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{0}\mathbf{.}\mathbf{380}\mathbf{}\mathbf{\text{s}}}\overrightarrow{\mathbf{v}}\mathit{d}\mathit{t}$

To perform the integral, you can use the calculus theorem

$\mathbf{\int }\left[A+\mathrm{Bf}\left(x\right)\right]\mathbf{dx}\mathbf{=}\mathbf{\int }\mathbf{A}\mathbf{dx}\mathbf{+}\mathbf{B}\mathbf{\int }\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx}$

You can think of the units and unit vectors as constants, represented by Aand B. Perform the integration to calculate the displacement of the ball from the edge of the table at 0.380 s.

Short answer

The displacement of the ball is$(0.456\text{m})\hat{\mathrm{i}}-(0.707\text{m})\hat{\mathrm{j}}$

Step-by-step solution

The given equation and theorem

The equation in the question at t = 0 is:

$\begin{array}{rcl}\mathrm{d}\overrightarrow{\mathrm{v}}& =& (1.2\text{m}/\text{s})\hat{\mathrm{i}}-(9.8\text{m}/\text{s})\mathrm{t}\hat{\mathrm{j}}\left(1\right)\\ \mathrm{\Delta }\overrightarrow{\mathrm{r}}& =& {\int }_0^{0.380\text{s}}\overrightarrow{\mathrm{v}}\text{d}\mathrm{t}\left(2\right)\end{array}$

During the time interval of 0.380 s

The calculus theorem is

$\int \left[\mathrm{A}+\mathrm{Bf}\left(\mathrm{x}\right)\right]\mathrm{dx}=\int \mathrm{A}\mathrm{dx}+\mathrm{B}\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

Concept

The rate of change of displacement with respect to time gives velocity. The area under the curve of velocity time graph gives the displacement in a given time interval.

The displacement of the ball

After substitution the value of$\overrightarrow{v}$in equation (2),the equation will be got that:

$\mathrm{\Delta }\overrightarrow{\mathrm{r}}={\int }_0^{0.380\mathrm{s}}\left[\left(1.2\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}-\left(9.8\mathrm{m}/\mathrm{s}\right)\mathrm{t}\hat{\mathrm{j}}\right]\mathrm{dt}\left(3\right)$

The calculus theorem which is given in the question is

$\int \left[\mathrm{A}+\mathrm{Bf}\left(\mathrm{x}\right)\right]\mathrm{dx}=\int \mathrm{A}\mathrm{dx}+\mathrm{B}\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

The theorem and the integration of equation (3) will be used :

$\mathrm{\Delta }\overrightarrow{\mathrm{r}}={\int }_0^{0.380\text{s}}(1.2\text{m/s})\hat{\mathrm{i}}\text{d}\mathrm{t}-(9.8\text{m/s})\hat{\mathrm{j}}{\int }_0^{0.380\text{s}}\mathrm{t}\text{d}\mathrm{t}$

After integration:

$\begin{array}{rcl}\mathrm{\Delta }\overrightarrow{\mathrm{r}}& =& (1.2\text{m/s})\hat{\mathrm{i}}\left[{\mathrm{t}}_0^{0.380\mathrm{s}}-\left(9.8\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}{\left[\frac{{\mathrm{t}}^2}{2}\right]}_0^{0.380\mathrm{s}}\right]\\ & =& \left(1.2\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}\left[\left(9.8\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}\right]\left[\frac{0.{380}^2}{2}-\frac{0^2}{2}\right]\\ & =& \left(0.456\mathrm{m}\right)\hat{\mathrm{i}}-\left(0.707\mathrm{m}\right)\hat{\mathrm{j}}\end{array}$

The displacement of the ball is $\begin{array}{rcl}& & \left(0.456\mathrm{m}\right)\hat{\mathrm{i}}-\left(0.707\mathrm{m}\right)\hat{\mathrm{j}}\end{array}$.