Question

The driver of the truck has an acceleration of 0.4g as the truck passes over the top A of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is 98 m, and the center of mass G of the driver (considered a particle) is 2 m above the road. Calculate the speed v of the truck.

Answer 1

19.81 m/s

Step-by-step explanation:

The total acceleration of the truck (a) is due to the centripetal acceleration and as a result of the linear acceleration. Therefore the total acceleration (a) is given by:

`a^2=a_n^2+a_t^2\\\\where\ a_n=centripetal\ acceleration=(v^2)/(r),a_t=linear \ acceleration\\\\But\ since\ the \ speed\ is \ constant, the \ linear \ acceleration(a_t)\ would\ be\ 0.\\\\a^2=a_n^2+a_t^2\\\\a^2=a_n^2\\\\a=a_n=(v^2)/(r) \\\\v^2=ar\\\\v=√(ar) \\\\a=0.4g=0.4*9.81,r=98\ m+2\ m=100\ m\\\\v=√(0.4*9.81*100) \\\\v=19.81\ m/s`