Question #226696

A car has an increasing speed at rate of 1.5m/s2 as it rounds a curve. What is the radius of curvature at a point when the speed of the car is 60kph and the magnitude of the total acceleration is 4m/s2?

Expert's answer

The velocity of the car =60kph=60∗1033600=16.67m/s= 60kph= 60* \frac{10^3}{3600}=16.67 m/s

The normal acceleration =an=v2R=16.672R=277.77R= a_n= \frac{v^2}{R}=\frac{16.67^2}{R}=\frac{277.77}{R}

Total acceleration

a=(aT)2+(an)24=(1.5)2+(277.77R)2  ⟹  R=75ma= \sqrt{(a_T)^2+(a_n)^2}\\ 4= \sqrt{(1.5)^2+(\frac{277.77}{R})^2}\\ \implies R = 75 m


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