Determine the stopping distance for an automobile going at a constant initial speed of 95 km/h and a human reaction time of 0.40 s (a) for acceleration$\mathbf{a} =-\mathbf{3}\mathbf{.}\mathbf{0} \mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$and (b) for$\mathbf{a} =-\mathbf{6}\mathbf{.}\mathbf{0} \mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$.

The acceleration of an object is defined as the rate at which the velocity of the object changes in a given amount of time.

Given data.

The initial speed of the car is $v_o=95\frac{\mathrm{km}}{\mathrm{h}}$.

The final speed of the car is$v=0$ as it comes to a stop.

Now,

$\begin{array}{c}{v}_o=95\frac{\mathrm{km}}{\mathrm{h}}\\ =95\frac{\mathrm{km}}{\mathrm{h}}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{1\mathrm{h}}{3600\mathrm{s}}\\ =26.39\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Rule.

If during the deceleration of the car, the traveled distance is $\left(x-x_o\right)$,

$\begin{array}{c}{v}^2=v_o^2+2a\left(x-x_o\right)\\ 2a\left(x-x_o\right)=v^2-v_o^2\\ x-x_o=\frac{v^2-v_o^2}{2a}\end{array}$

Then,

role="math" localid="1642839079982" $x=x_o+\frac{v^2-v_o^2}{2a}\dots \left(i\right)$.

Here, the car moves at the same speed $v_o$during the reaction time before applying the brakes.

The reaction time is $t_{\mathrm{r}} = 0.40 \mathrm{s}$.

Therefore, due to the reaction time, the car travels a distance of $x_o = v_o t_{\mathrm{r}}$before the application of the brakes.

Now,

$\begin{array}{c}{x}_o = 26.39 \frac{\mathrm{m}}{\mathrm{s}} \times 0.40 \mathrm{s}\\ =10.56 \mathrm{m}\end{array}$

(a) When the acceleration is $a=-3.0\frac{\mathrm{m}}{{\mathrm{s}}^2}$, from equation (i),

$\begin{array}{c}x=10.56\mathrm{m}+\frac{0^2-{\left(26.39\frac{\mathrm{m}}{\mathrm{s}}\right)}^2}{2\left(-3.0\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}\\ =126.63\mathrm{m}\end{array}$

Therefore, the stopping distance is 126.63 m when the acceleration is $a=-3.0\frac{\mathrm{m}}{{\mathrm{s}}^2}$.

(b) When the acceleration is $a=-6.0\frac{\mathrm{m}}{{\mathrm{s}}^2}$, from equation (i),

$\begin{array}{c}x=10.56\mathrm{m}+\frac{0^2-{\left(26.39\frac{\mathrm{m}}{\mathrm{s}}\right)}^2}{2\left(-6.0\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}\\ =68.60\mathrm{m}\end{array}$

Therefore, the stopping distance is 68.60 m when the acceleration is $a=-6.0\frac{\mathrm{m}}{{\mathrm{s}}^2}$.