91Ó°ÊÓ

The force acting on the object is varying with time so the distance covered by an object in a given duration can be found by integrating the force twice.

Given Data:

The mass of the object is $\mathrm{m}=45\mathrm{kg}$.

The magnitude of the force is $\mathrm{F}\left(\mathrm{t}\right)=\left(16.8\mathrm{N}/\mathrm{s}\right)\mathrm{t}$.

The duration of travelled distance by an object is $\mathrm{t}=5\mathrm{s}$.

The distance travelled by object is given as:

$$\begin{gathered} \mathrm{F}\left(\mathrm{t}\right)=16.8\mathrm{t} \\ \mathrm{ma}=16.8\mathrm{t} \\ \mathrm{m}\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)=16.8\mathrm{t} \\ \mathrm{dv}=\left(\frac{16.8}{\mathrm{m}}\right)\mathrm{dt} \end{gathered}$$

Take an integration both sides.

$$\begin{gathered} \underset{0}{\overset{\mathrm{v}}{∫}}\mathrm{dv}=\left(\frac{16.8}{\mathrm{m}}\right)\underset{0}{\overset{\mathrm{t}}{∫}}\mathrm{dt} \\ \mathrm{v}-0=\left(\frac{16.8}{\mathrm{m}}\right)\left(\frac{{\mathrm{t}}^2}{2}-0\right) \\ \frac{\mathrm{dx}}{\mathrm{dt}}=\left(\frac{8.4}{\mathrm{m}}\right){\mathrm{t}}^2 \\ \mathrm{dx}=\left(\frac{8.4}{\mathrm{m}}\right){\mathrm{t}}^2\mathrm{dt} \end{gathered}$$

Again take an integration both sides.

$$\begin{gathered} \underset{0}{\overset{\mathrm{d}}{∫}}\mathrm{dx}=\left(\frac{8.4}{\mathrm{m}}\right)\underset{0}{\overset{\mathrm{t}}{∫}}{\mathrm{t}}^2\mathrm{dt} \\ \mathrm{d}-0=\left(\frac{8.4}{\mathrm{m}}\right)\left(\frac{{\mathrm{t}}^3}{3}-0\right) \\ \mathrm{d}=\left(\frac{8.4}{\mathrm{m}}\right){\mathrm{t}}^3 \end{gathered}$$

Here, $\mathrm{m}$is the mass of object, $\mathrm{t}$is the duration for travel, $\mathrm{d}$distance travelled by object.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{d}=\left(\frac{8.4}{3\left(45\mathrm{kg}\right)}\mathrm{N}/\mathrm{s}\right){\left(5\mathrm{s}\right)}^3 \\ =7.78\mathrm{m} \end{gathered}$$

Therefore, the distance travelled by object is $7.78\mathrm{m}$.