A car that weighs $\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}\mathbf{}}\mathbf{N}$ is initially moving at role="math" localid="1656996519190" $\mathbf{40}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}$ when the brakes are applied and the car is brought to a stop in role="math" localid="1656996543045" $\mathbf{15}\mathbf{}\mathbf{m}$. Assuming the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

• Mass of the car is$\mathrm{m}=1.30\times {10}^4\mathrm{N}$ ,
• The initial velocity of the car is${\mathrm{v}}_0=40\frac{\mathrm{km}}{\mathrm{hr}}\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\left(\frac{1\mathrm{hr}}{3600\mathrm{s}}\right)=11.2\mathrm{m}/\mathrm{s}$ ,
• The stopping distance is$∆\times =15\mathrm{m}$ .

Kinematic equations are a set of equations of motion that relate to the velocity, acceleration, displacement, and time of the motion of a particle.

We have given the weight of the car, so we can calculate its mass. We can calculate the car’s acceleration as we have its speed and stopping distance. As we know the acceleration so, we can calculate the force on the car.

Formulae.

${\mathbf{v}}^{\mathbf{2}}\mathbf{}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{∆}\mathbf{\times }$ (1)

$\mathbf{v}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{v}}_{\mathbf{0}\mathbf{}}\mathbf{+}\mathbf{}\mathbf{at}$ (2)

localid="1657002300051" $\mathbf{F}\mathbf{}\mathbf{=}\mathbf{ma}$ (3)

We can find the mass of the body from the equation of weight.

$\mathrm{W}=\mathrm{mg}$

$\mathrm{m}=\frac{\mathrm{W}}{\mathrm{g}}$

$=\frac{1.30\times {10}^4\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2}$

$=1326\mathrm{kg}$

Therefore, the mass of the car is$1326\mathrm{kg}$ .

We know that the final velocity of the car is zero. By using the given initial velocity and stopping distance in equation (1), we can find the acceleration of the car.

Rearrange the equation (1) and substitute the value of the given quantities, we get

$\mathrm{a}=-\frac{{\mathrm{v}}_0^2}{2∆\times }$

$$\begin{gathered} =-{\frac{\left(11.2\mathrm{m}/\mathrm{s}\right)}{2\times 15\mathrm{m}}}^2 \\ =-4.12\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the magnitude of the acceleration is $-4.12\mathrm{m}/{\mathrm{s}}^2$

We know that force is a stopping force that acts in the opposite direction of motion. Therefore, equation (3) can be written as,

$$\begin{gathered} -\mathrm{F}=\mathrm{ma} \\ -\mathrm{F}=1327\mathrm{kg}\times \left(-4.12\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{F}=5.5\times {10}^3\mathrm{N} \end{gathered}$$

Therefore, the stopping force is equal to $5.5\times {10}^3\mathrm{N}$ .

The final velocity is zero as the car comes to stop. We have calculated acceleration in the above part. Using the value of initial, final velocity, and acceleration in equation (2), we can find the time required for stopping the car. Substitute the values in equation (2).

$$\begin{gathered} \mathrm{t}=-\frac{{\mathrm{v}}_0}{\mathrm{a}} \\ =-\left(\frac{11.11\mathrm{m}/\mathrm{s}}{-4.12\mathrm{m}/{\mathrm{s}}^2}\right) \\ =2.7\mathrm{s} \end{gathered}$$

Therefore, the time required for the car to stop is $2.7\mathrm{s}$

If we doubled the initial speed of the car then we know that from the kinematic equation we can say that ${\mathrm{v}}_0$are $∆\times$directly proportional to each other. So, doubling of ${\mathrm{v}}_0$is a quadrupling of $∆\times$. That is, it increases by factor 4.

From equation (2), we can see that $\mathrm{t}\mathrm{and}{\mathrm{v}}_0$are directly proportional. So, doubling one is doubling the other. Therefore, if we double the ${\mathrm{v}}_0$ then stopping time will increase by a factor of 2.