Question

The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the speed . if 210 lb of force keeps a 2,500 lb car from skidding on a curve of radius 500 ft at 25 mph . what force would keep the same car from skidding on a curve of radius 750 ft at 40 mph

Answer 1

@jim_thompson5910

Answer 2

what is the initial equation? were you able to figure that out?

Answer 3

no i'm lost on this

Answer 4

i really forgot how to do this one

Answer 5

ok we are going to have these variables

F = force needed to keep a car from skidding on a curve
r = radius
w = weight of the car
s = speed of the car

Answer 6

ok

Answer 7

"The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve"

that allows us to say

F = k/r

Answer 8

tack on "jointly as the weight of the car and the square of the speed" and you'll have

\[\Large F = \frac{k*w*s^2}{r}\]

where k is some constant

Answer 9

ok

Answer 10

now we get to the part " 210 lb of force keeps a 2,500 lb car from skidding on a curve of radius 500 ft at 25 mph"


F = 210
w = 2500
r = 500
s = 25

plug those values into \[\Large F = \frac{k*w*s^2}{r}\] and solve for k

Answer 11

ok

Answer 12

k=42/625

Answer 13

its looking for lb

Answer 14

k = 42/625 = 0.0672

good

Answer 15

"what force would keep the same car from skidding on a curve of radius 750 ft at 40 mph"

so it wants the value of F where

k = 0.0672
w = 2500
r = 750
s = 40

Answer 16

210

Answer 17

that was the old value of F

Answer 18

750

Answer 19

plug

k = 0.0672
w = 2500
r = 750
s = 40

into

\[\Large F = \frac{k*w*s^2}{r}\]

Answer 20

268.800

Answer 21

oops 537.6.

Answer 22

\[\Large F = \frac{k*w*s^2}{r}\]

\[\Large F = \frac{0.0672*2500*40^2}{750}\]

\[\Large F = \frac{268,800}{750}\]

\[\Large F = 358.4\]

Answer 23

oh okay

Answer 24

i finally caught on how this was done Thank you