The resistance(R) to the motion of a car is partly constant and partly varies as the square of the speed, u. When the car is moving at 30kg per hour the resistance is 630N and when the car is moving at 50kg per hour the resistance is 950N
I. Find an expression for R in terms of u
II. The resistance at 80kg per hour
R = au^2 + b
You have two data points, so just solve for a and b in
900a+b = 630
2500a+b = 950
Then find R(80)
oh -- and speed is km/hr, not kg/hr
I. To find the expression for R in terms of u, we can set up a system of equations using the given information.
Let R₀ be the constant resistance and Rₚ be the resistance proportional to the square of the speed.
We know that when the car is moving at 30 km per hour, the resistance is 630N. So, we have the equation:
R₀ + Rₚ = 630 ----(Eq. 1)
We also know that when the car is moving at 50 km per hour, the resistance is 950N. So, we have the equation:
R₀ + 2500Rₚ = 950 ----(Eq. 2)
To find the values of R₀ and Rₚ, we can solve this system of equations.
Subtracting Eq. 1 from Eq. 2, we get:
2500Rₚ - Rₚ = 950 - 630
2499Rₚ = 320
Rₚ = 320/2499 ≈ 0.128
Now, substituting the value of Rₚ into Eq. 1, we can find R₀:
R₀ + 0.128 = 630
R₀ = 630 - 0.128
R₀ ≈ 629.872
Therefore, the expression for R in terms of u is:
R = 629.872 + 0.128u
II. To find the resistance at 80 km per hour, we can substitute u = 80 into the expression for R:
R = 629.872 + 0.128(80)
R ≈ 629.872 + 10.24
R ≈ 640.112
So, the resistance at 80 km per hour is approximately 640.112N.
To find an expression for R in terms of u, we can use the given information that the resistance is partly constant and partly varies as the square of the speed, u.
Let's denote the constant resistance as K. We can express the resistance as the sum of the constant resistance and the variable resistance:
R = K + a(u^2)
where K is the constant resistance, a is the coefficient of variation, and u is the speed.
Now, we need to determine the values of K and a using the given information.
At 30 kg per hour, the resistance is given as 630N. Substituting these values into the equation:
630 = K + a(30^2)
At 50 kg per hour, the resistance is given as 950N. Substituting these values into the equation:
950 = K + a(50^2)
Now, we have a system of two equations with two unknowns (K and a). We can solve this system to find the values of K and a.
Let's simplify the equations:
630 = K + 900a --> Equation 1
950 = K + 2500a --> Equation 2
We can solve this system by subtracting Equation 1 from Equation 2:
950 - 630 = K + 2500a - K - 900a
320 = 1600a
Divide both sides by 1600:
a = 0.2
Now, substitute this value of a back into Equation 1 or Equation 2 to find the value of K:
630 = K + 900(0.2)
630 = K + 180
Subtract 180 from both sides:
450 = K
Now we have found the values of K and a:
K = 450
a = 0.2
Therefore, the expression for R in terms of u is:
R = 450 + 0.2(u^2)
To find the resistance at 80 kg per hour, we can substitute u = 80 into the expression for R:
R = 450 + 0.2(80^2)
R = 450 + 0.2(6400)
R = 450 + 1280
R = 1730 N
Therefore, the resistance at 80 kg per hour is 1730 N.