91Ó°ÊÓ

The given data can be listed below,

• The weight of the driver is,$m_d=150\text{ l²ú}\left(0.45\text{ k²µ/1 l²ú}\right)=67.5\text{ k²µ}$
• The initial time of the car is$t=0$
• The final time of the car is$t_2=4.50\text{ s}$,
• The net braking force of the car is,$B_x=−5200\text{ N}$

In addition to being the result of force and time, momentum change is another way to characterise impulse.

The impulse of the car in a time interval is given by,

$J_X={∫}_{t_1}^{t_2}{F}_x\left(t\right)dx$

Which is also the area under the curve of the graph which can be calculated as,

$\begin{array}{c}{J}_x=(7500\left(1.5\right)+\left(0.5\right)(7500+3500)(3−15)+3500×(4.5−3))\text{ N}â‹\dots \text{s}\\ =24750\text{ N}â‹\dots \text{s}\end{array}$

Thus, the impulse of the car I a time interval 0 to 4.50 s is $24750\text{ N}â‹\dots \text{s}$.

The impulse is also known as change in momentum which can be given by,

$\begin{array}{c}{J}_x=p_x\\ =m{v}_x\\ v_x=\frac{J_x}{m}\end{array}$

Here, m is the mass of the car and $J_x$is the impulse of the car.

Substitute all the values in the above,

$\begin{array}{c}{v}_x=\frac{J_x}{m}\\ =\frac{24750\text{ N}â‹\dots \text{s}}{3071\text{ l²ú}\left(\frac{0.454\text{ k²µ}}{1\text{ l²ú}}\right)}\left(\frac{1\text{ k²µ}â‹\dots \text{m/s}}{1\text{ N}}\right)\\ =17.75\text{″¾/²õ}\end{array}$

Thus, the speed of the car at 4.50 s is$17.75\text{″¾/²õ}$ .

The time taken to stop the car can be determined by impulse applied by the braking force which is given by,

$\begin{array}{c}{J}_x=B_x{t}_b\\ t_b=\frac{J_x}{B_x}\end{array}$

Here,$J_x$ is the impulse of the car and $B_x$is the net braking force.

Substitute all the values in the above,

$\begin{array}{c}{t}_b=\frac{24750\text{ N}â‹\dots \text{s}}{−5200\text{ N}}\\ =4.76\text{ s}\end{array}$

Thus, the time at which the car will stop after applying braking force is$4.76\text{ s}$ .

The work done by the braking force is the change in the kinetic energy of the car, as the final kinetic energy of the car is zero the work done is given by,

$W_{B_x}=K_f−K_{in}$

Substitute all the values in the above,

$\begin{array}{c}{W}_{B_x}=0−\frac{1}{2}m{v}_x^2\\ =−\frac{1}{2}(3071×0.454\text{ k²µ}){\left(17.75\text{″¾/²õ}\right)}^2\\ =−2.20×{10}^5\text{ J}\end{array}$

Thus, the work done by the braking force to stop the car is $−2.20×{10}^5\text{ J}$.

From the work done applied by braking force to stop the car the distance travelled by the car can be determined by,

$\begin{array}{c}{W}_{B_x}=B_{x}l\\ l=\frac{W_{B_x}}{B_x}\end{array}$

Here,$W_{B_x}$ is the work done by the braking force and $B_x$is the braking force on the car.

Substitute all the values in the above,

$\begin{array}{c}l=\frac{−2.20×{10}^5\text{ J}}{−5200\text{ N}}\\ =42.3\text{″¾}\end{array}$

Thus, the distance travelled by the car before it stops is$42.3\text{″¾}$ .