The resistance QR to the motion of a car is partly constant and varies as the square of the speed. At 20km/hr it is 265N. And at 30km/hr it is 365N. Find the resistance at 80km/hr
QR = a s^2 + c
265N = a 20^2 + c = 400 a + c
365N = a 30^2 + c = 900 a + c
subtracting equations (to eliminate c) ... 500 a = 100N
solve for a , then substitute back to find c
use the new equation to find QR at 80km/hr
oops ... my bad. I missed the "square" part.
But I'm sure you caught it before R_Scott did ... right?
To find the resistance at 80 km/hr, we can use the given information to create a relationship between the speed and resistance.
Let's assume that the constant resistance is represented by "a" N, and the variable resistance (which varies as the square of the speed) is represented by "b" N/(km/hr)^2.
We are given the following two equations:
at 20 km/hr: a + b(20)^2 = 265 N
at 30 km/hr: a + b(30)^2 = 365 N
To find the values of "a" and "b", we can use these two equations to form a system of linear equations.
From the first equation, we have:
a + 400b = 265 (equation 1)
From the second equation, we have:
a + 900b = 365 (equation 2)
To solve this system of linear equations, we can subtract equation 1 from equation 2:
(a + 900b) - (a + 400b) = 365 - 265
500b = 100
b = 0.2 N/(km/hr)^2
Now that we have the value of "b", we can substitute it into one of the original equations to find the value of "a". Let's use equation 1:
a + 400(0.2) = 265
a + 80 = 265
a = 265 - 80
a = 185 N
So, we have found the values of "a" and "b" to be a = 185 N and b = 0.2 N/(km/hr)^2.
To find the resistance at 80 km/hr, we can substitute the speed into the variable resistance equation:
Resistance at 80 km/hr = a + b(80)^2
Resistance at 80 km/hr = 185 + 0.2(80)^2
Resistance at 80 km/hr = 185 + 0.2(6400)
Resistance at 80 km/hr = 185 + 1280
Resistance at 80 km/hr = 1465 N
Therefore, the resistance at 80 km/hr is 1465 N.
R = a + bs
to find a and b, solve
a+20b = 265
a+30b = 365
then find a+80b