Physics
posted by Hillary on .
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 onefourth of the circle.

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 = (v₂v₁)/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