A 5.0 kg toy car can move along an x-axis; Figure 9-50 gives${\mathit{F}}_{\mathbf{x}}$of the force acting on the car, which begins at rest at time t = 0 . The scale on the ${\mathit{F}}_{\mathbf{x}}$axis is set by${\mathbf{F}}_{\mathbf{xs}\mathbf{}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$ . In unit-vector notation, what is p at (a) t = 4.0 sand (b) t = 7.0 s, and (c) what is at t = 9.0 s?

The mass of the toy car m = 5.0 kg .

The ${\mathrm{F}}_{\mathrm{xs}}=5.0\mathrm{N}$.

The initial speed of the car is 0 m/s .

The start of motion of the car,${\mathrm{t}}_0=0_{\mathrm{s}}$ .

The graphical presentation of force and time gives another method to determine the momentum or force in a given situation. The area under the curve in a force-time graph gives us the momentum of the particle in the motion which is given below.

$\begin{array}{l}\overrightarrow{\mathrm{J}}=∆\overrightarrow{\mathrm{p}}\\ =\mathrm{m}∆\overrightarrow{\mathrm{v}}\\ \mathrm{J}={\mathrm{F}}_{\mathrm{avg}}\end{array}∆\mathrm{t}$

The impulse–linear momentum theorem gives

$$\begin{gathered} \overrightarrow{\mathrm{J}}=∆\overrightarrow{\mathrm{p}} \\ ={\mathrm{F}}_{\mathrm{avg}}∆\mathrm{t} \end{gathered}$$

From the graph given in figure 9-50, we can see that,

$$\begin{gathered} ∆\overrightarrow{\mathrm{p}}=\mathrm{area}\mathrm{under}\mathrm{the}\mathrm{curve}\mathrm{from}\mathrm{t}=0\mathrm{to}\mathrm{t}=4.0\mathrm{s} \\ =\mathrm{area}\mathrm{of}\mathrm{triangle}+\mathrm{area}\mathrm{of}\mathrm{rectangle} \\ =\left(\frac{1}{2}\times \mathrm{b}\times \mathrm{h}\right)+\left(\mathrm{l}\times \mathrm{w}\right) \\ =\left(\frac{1}{2}\times 2\mathrm{s}\times 10\mathrm{N}\right)+\left(10\mathrm{N}\times 2\mathrm{s}\right) \\ =30\mathrm{kg}.\mathrm{m}/\mathrm{s}\hat{\mathrm{i}} \end{gathered}$$

The impulse-linear momentum value is 30 kg.$\mathrm{m}/\mathrm{s}\hat{\mathrm{i}}$

Similar to part (a), from the graph we can see that

$$\begin{gathered} ∆\overrightarrow{\mathrm{p}}=\mathrm{area}\mathrm{under}\mathrm{the}\mathrm{curve}\mathrm{from}\mathrm{t}=0\mathrm{to}7.0\mathrm{s} \\ =\mathrm{area}\mathrm{of}\mathrm{trapezoid}+\mathrm{area}\mathrm{of}\mathrm{triangle} \\ ∆\overrightarrow{\mathrm{p}}=\left(\frac{1}{2}\times (\mathrm{a}+\mathrm{b})\times \mathrm{h}\right)+\left(\frac{1}{2}\times \mathrm{b}\times \mathrm{h}\right) \\ =\left(\frac{1}{2}\times (2\mathrm{s}+6\mathrm{s})\times 10\mathrm{N}\right)+\left(\frac{1}{2}\times 1\mathrm{s}\times \left(-5.0\mathrm{N}\right)\right) \\ =40\mathrm{kg}.\mathrm{m}/\mathrm{s}-2.5\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ =37.5\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Similar to part(a), from the graph we will calculate the momentum and then velocity as,

$$\begin{gathered} ∆\overrightarrow{\mathrm{p}}=\mathrm{area}\mathrm{under}\mathrm{the}\mathrm{curve}\mathrm{from}\mathrm{t}=0\mathrm{to}9.0\mathrm{s} \\ =\mathrm{area}\mathrm{of}\mathrm{trapezoid}+\mathrm{area}\mathrm{of}\mathrm{small}\mathrm{trapezoid} \\ ∆\overrightarrow{\mathrm{p}}=\left(\frac{1}{2}\times ({\mathrm{a}}_1+{\mathrm{b}}_1)\times {\mathrm{h}}_1\right)+\left(\frac{1}{2}\times \mathrm{b}\times \mathrm{h}\right) \\ =\left(\frac{1}{2}\times (2\mathrm{s}+6\mathrm{s})\times 10\mathrm{N}\right)+\left(\frac{1}{2}\times \left(1\mathrm{s}+3\mathrm{s}\right)\times \left(-5.0\mathrm{N}\right)\right) \\ =40\mathrm{N}.\mathrm{s}-10\mathrm{N}.\mathrm{s} \\ =30\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ =30\mathrm{kg}.\mathrm{m}/\mathrm{s}\hat{\mathrm{i}} \end{gathered}$$

Now the velocity can be determined as

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{P}}{\mathrm{m}} \\ \mathrm{v}=\frac{30\mathrm{kg}.\mathrm{m}/\mathrm{s}}{5.0\mathrm{kg}} \\ \mathrm{v}=6.0\mathrm{m}/\mathrm{s} \\ \overrightarrow{\mathrm{v}}=6.0\mathrm{m}/\mathrm{s}\hat{\mathrm{i}} \end{gathered}$$