Change in velocity
Δv⃗=v⃗₂-v⃗₁
|Δv|=sqrt(v²+v²-2v•v•cos60°)=sqrt(2v²(1-cos60°))=
=sqrt{2•60(1-0.5)}=7.74 m/s
The distance covered by the car is
s= 2•π•R•60°/360°=2•π•300/6=314 m
t= s/v=314/60=5.23 s
a=|Δv| /t=7.74/5.23 =1.48 m/s²
Centripetal acceleration (instantaneous acceleration)
a= v²/R= 60²/300=12 m/s²
CIRCULAR MOTION
A car takes a curve with 300m radius at a velocity of 60 m/s.
a) Determine the change in velocity (module and direction) when it travels an arc of 60 degrees
b)Compare the magnitude of the instantaneous acceleration of the car with a magnitude of the average acceleration to transverse the arc
3 answers
Change in velocity
vecorΔv =vecor v₂-vecorv₁
vecorΔv =vecor v₂-vecorv₁
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