Question

3. A car is traveling up a 3% grade, with the speed of 85mph, on a road that has good, wet pavement. A deer jumps out onto the road and the driver applies the brakes 290-ft from it. The driver hits the deer at a speed of 20mph.If the driver did not have antilock brakes, and the wheels were locked the entire distance, would a deer-impact speed of 20mph be possible

Answer 1

Answer:

Explanation:

From the given information;

Let assume that:

the wheel radius = 15 inches

the driveline slippage = 3%;  &

the gear reduction ratio (overall) = 2.5 to 1

So; using the equation:

$v_1= \dfrac{2 \pi r n_o (1 -i)}{\varepsilon_o}$

$v_1= \dfrac{2 \times 3.14 \times \dfrac{15}{12} \times \dfrac{85}{100} (1 -0.03)}{2.5}$

$v_1= \dfrac{2 \times 3.14 \times \dfrac{15}{12} \times \dfrac{85}{100} (0.97)}{2.5}$

$v_1 = 126.92 \ fp^3$

$frl = 0.01 ( 1+ \dfrac{v}{147}) \ if \ v \ is \ ft/sec$

$frl = 0.01 \Bigg( 1+ \dfrac{\dfrac{126.92 +(20)1.47 }{2} }{147}\Bigg)$

$S = \dfrac{v_b ( v_1^r-v_2^r)}{2g(n_b \mu + frl \pm sin \ y}$

where;

$\mu = 0.6$

$291 = \dfrac{1.64( 126.92^2-29.9^2)}{64.4(n_b \times 0.6 +0.01532 +0.03}$

$n_b = 1.33 \to which \ is \ not \ possible$

However;

$n_b \mu = 1.33(0.6) = 0.80$

$\mu = 0.9 \to$ if the car's anti-clock breaking system did not fail

Thus;

$n_b (0.9) = 0.80$

$n_b =\dfrac{ 0.80}{(0.9) }$

$n_b = 0.89$

Hence, the distance is possible if the anti-clock breaking system did not fail.