Dear Ram, so here it goes-
Speed of car on the road , v = 36km/h
= >36 × 5/18 = 10m/s
coefficient of friction between tires and road =0.5
Let radius of car is r
use formula, v=squareroot(u*r*g)
where g is acceleration due to gravity and u is coefficient of friction.
10×10 = 0.5 × r × 10 [taken g = 10m/s² ]
r = 20m
hence, minimum turning radius of car is 20m.
All the best
Hope this helps
Question : The force (in pound-force) needed to keep a car from skidding on a curve varies directly with the weight of the car (in pounds) and the square of its speed (in miles per hour [mph]) and inversely with the radius (in feet) of the curve. Suppose 6125 pound-force is required to keep a 2750 pound car, travelling at a speed of 35 mph, from skidding on a curve of radius 550 feet. How much pound-force is then required to keep a 3600-pound car, travelling at a speed of 50 mph, from skidding on a curve of radius 750 feet?
Option 1: 11960
Option 2: 12150
Option 3: 12240
Option 4: 12000
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