The driver of a truck has an acceleration of 0.3 g as the truck passes over the top of a hump in the road at a constant speed. The radius of curvature of the road at the top of the road is 98 m, and the center of mass G of the driver, who can be considered a particle, is 2 m above the road. What is the speed of the truck

17.115m/s

Explanation:

The formula that will be used to calculate the speed of the truck is expressed as shown:

v = √ar

a is the acceleration of the truck

r is the radius of curvature of the road at the top of the road plus centre of mass G who can be considered a particle, is 2 m

v = √a(R+G)

Given parameters

a = 0.3g (g is the acceleration due to gravity)

R = 98m

G = 2m

Substituting the values into the formula:

v = √0.3(9.81){98+2}

v = √0.3×9.81×100

v = √294.3

v = 17.155m/s

Hence the speed of the truck is 17.155m/s

12000j is work done

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement