Posted by **Hillary** on Monday, January 21, 2013 at 11:35pm.

A car rounds a 1.35km radius circular track at 93 km/h.

Find the magnitude of the car's average acceleration after it has completed one-fourth of the circle.

- Physics -
**Elena**, Tuesday, January 22, 2013 at 9:56am
93 km/h=93000/3600=25.8 m/s

Centripetal acceleration is

a=v²/R=25.8²/1350 =0.49 m/s²

The average acceleration =

=Δv/Δt = (v₂-v₁)/Δt

Since velocity is vector, vecor Δv can be found as vector difference (taking into account that velocity vector changed its direction by 90 degrees after the point has completed ¼ of the circle) => Δv = v•√2.

The time for the uniform motion for ¼ of the circle is

Δt=s/v=(2πR)/4v =π•1350/2•25.8 = 82.2 s,

a(ave) = Δv/Δt = 25.8•√2/82.2 =

=0.44 m/s²

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