Respuesta :
Given:
Stopping distance range is d = (65, 70) ft.
The stopping distance, d, obeys this formula.
d = v²/(2μg)
where
v = speed of the vehicle
μ = 0.8, coefficient of static friction under good road conditions
g = acceleration due to gravity, 32.2 ft/s²
Therefore
v = √(2*0.8*32.2*d) = 7.178√d
Test d = 65 ft.
v = 7.178√(65) = 57.87 ft/s = (57.87/88)*60 = 39.5 mph
Test d = 70 ft.
v = 7.178√(70) = 60.05 ft/s = 40.9 mph
To be safe, the lower speed of 39.5 mph is preferred.
Answer: 40 mph
Stopping distance range is d = (65, 70) ft.
The stopping distance, d, obeys this formula.
d = v²/(2μg)
where
v = speed of the vehicle
μ = 0.8, coefficient of static friction under good road conditions
g = acceleration due to gravity, 32.2 ft/s²
Therefore
v = √(2*0.8*32.2*d) = 7.178√d
Test d = 65 ft.
v = 7.178√(65) = 57.87 ft/s = (57.87/88)*60 = 39.5 mph
Test d = 70 ft.
v = 7.178√(70) = 60.05 ft/s = 40.9 mph
To be safe, the lower speed of 39.5 mph is preferred.
Answer: 40 mph
The highest speed from which you could stop the vehical with good brakes before hitting him is 40 mph. Therefore the correct option ia c) 40mph.
Given :
Stopping Distance range - (65ft , 70ft)
Solution :
We know that the stopping distance (d) is,
[tex]\rm d = \dfrac{v^2}{2\mu g }[/tex] --- (1)
Where,
v = speed of the vehicle.
[tex]\mu[/tex] = 0.8, coefficient of static friction under good road conditions.
--- (acceleration due to gravity)
Now at d = 65 ft
put the values of v, g and [tex]\mu[/tex] in equation (1) we get,
[tex]\rm v = \sqrt{2\mu g d}[/tex]
[tex]\rm v = \sqrt{2\times 0.8 \times 32.2 \times 65}[/tex]
v = 57.87 ft/sec = 39.5 mph
Now at d = 70 ft
Put the values of v, g and [tex]\mu[/tex] in equation (1) we get,
[tex]\rm v = \sqrt{2\times 0.8 \times 32.2 \times 70}[/tex]
v = 60.05 ft/sec = 40.9 mph
So the safe speed is 39.5 mph approximately 40 mph. Therefore the correct option is c) 40 mph.
For more information, refer the link given below
https://brainly.com/question/4726243
