The algebraic relation d= 0.0056s2 +0.14s models the relation between a vehicle's stopping distance d, in metres and its speed s, in km/h.
a) What is the fastest you could drive and still be able to stop within 80m?
b) What is the stopping distance for a car traveling at 120 km/h?
Online "^" is used to indicate and exponent, e.g., x^2 = x squared. I assume only the s value is squared.
a. 80 = 0.0056(s^2) +0.14s
b. d= 0.0056(120^2) +0.14(120)
Solve for the unknown.
a) To find the maximum speed that still allows the vehicle to stop within 80m, we need to solve the equation for s when d is equal to 80.
0.0056s^2 + 0.14s = 80
Subtracting 80 from both sides of the equation gives:
0.0056s^2 + 0.14s - 80 = 0
This is a quadratic equation in standard form. To solve it, we can use the quadratic formula:
s = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values of a, b, and c from the equation gives:
s = (-(0.14) ± √((0.14)^2 - 4(0.0056)(-80))) / (2(0.0056))
Simplifying further:
s = (-0.14 ± √(0.0196 + 0.1792)) / 0.0112
s = (-0.14 ± √0.1988) / 0.0112
s = (-0.14 ± 0.4456) / 0.0112
s ≈ (-0.5856) / 0.0112 or (0.3056) / 0.0112
s ≈ -52.32 or 27.32
Since speed cannot be negative, the maximum speed that still allows the vehicle to stop within 80m is approximately 27.32 km/h.
b) To find the stopping distance for a car traveling at 120 km/h, we can plug in s = 120 into the original equation:
d = 0.0056(120)^2 + 0.14(120)
Simplifying:
d = 0.0056(14400) + 16.8
d = 80.64 + 16.8
d ≈ 97.44 m
The stopping distance for a car traveling at 120 km/h is approximately 97.44 meters.
To find the answers to these questions, we need to use the given algebraic relation: d = 0.0056s^2 + 0.14s, where d is the stopping distance in meters and s is the speed in km/h.
a) We want to find the fastest speed at which we can still stop within 80m. To do this, we need to solve the equation for s when d is equal to 80.
Setting d = 80 in the equation, we get:
80 = 0.0056s^2 + 0.14s
To solve this quadratic equation, we can either factorize it or use the quadratic formula. Let's use the quadratic formula:
The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Applying this formula to our equation, we have:
s = (-0.14 ± √(0.14^2 - 4(0.0056)(-80))) / (2(0.0056))
Calculating the values inside the square root, we get:
s = (-0.14 ± √(0.0196 + 1.792)) / 0.0112
s = (-0.14 ± √1.8116) / 0.0112
Taking the square root, we get:
s = (-0.14 ± 1.346) / 0.0112
Now, we have two possible values for s:
s = (-0.14 + 1.346) / 0.0112 ≈ 110.18 km/h
s = (-0.14 - 1.346) / 0.0112 ≈ -129.46 km/h
Since we cannot have negative speed in this context, the fastest we can drive and still be able to stop within 80m is approximately 110.18 km/h.
b) To find the stopping distance for a car traveling at 120 km/h, we need to plug the speed value into the given equation.
Using s = 120 km/h in the equation, we have:
d = 0.0056(120)^2 + 0.14(120)
Calculating the values, we get:
d = 0.0056(14400) + 16.8
d = 80.64 + 16.8
d = 97.44 meters
Therefore, the stopping distance for a car traveling at 120 km/h is approximately 97.44 meters.