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A truck increases its speed at constant ...

A truck increases its speed at constant rate of `0.1m//s^(2)` and it passes over a semicircular bridge of radius 40 m. At the time car reaches the top of the bridge, its speed is 2m/s. What are the magnitude and direction of the to total acceleration vector at this moment?

Text Solution

Verified by Experts

The truck moves long a curved path with variable speed. Thus, the truck has both tangential and radial acceleration. The radial acceleration is given by `a_(r)=v^(2)//R`, with `v=2m//s` and R=40m. The radial acceleration vector is directed straight downward, and the tangential acceleration vector has magnitude `0.1m//s^(2)` and its is horizontal.
Calculation: Let us calculate the radial acceleration:
`a_(r)=(v^(2))/R=((2m//s)^(2))/(40m)=0.1m//s`
Now
`|bara|=sqrt(a_(r)^(2)+a_(4)^(2))=sqrt(2)(0.1)m//s^(2)`
Which is the magnitude of `veca`. Now, the angle `phi` between `veca` and the horizontal is given by
`phi=tan^(-1)((a_(r))/(a_(t)))=45^(@)`
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