Question

A car traveling at a speed of v can brake to an emergency stop in a distance x , Assuming all other driving conditions are all similar , if the traveling speed of the car double, the stopping distance will be (1) √2x,(2)2x, or(3) 4x:(b) A driver traveling at 40.0km/h in school zone can brake to an emergency stop in 3.00m. What would be braking distance if the car were traveling at 60.0 km/h?

Answer 1

D = 6.74 m

Step-by-step explanation:

Let the f newtons be the braking force

distance to stop = x meters

speed = v m/s

we know that work done is given as

work done W = fx joules

`fx = (1)/(2) mv^2`

If the speed is doubled ,

`fx' = (1)/(2) m(2v)^2`

`= 4[(1)/(2)mv^2] = 4fx`

stooping distance is D = 4x

`40km/h = (40*1000)/(60*60) = 11.11 m/s`

60 km/hr = 16.66 m/s

braking force = f

`f*3 =  [(1)/(2)mv^2] = [(1)/(2)m*11.11^2]`

`f*D = [(1)/(2)m*16.66^2]`

`(D)/(3) =  (16.66^2)/(11.11^2)`

`(D)/(3) = 2.2489`

D = 6.74 m