A particle moves along a circle with speed v=at . The total acceleration of the point at a time when it has traced 1/8th of the circumference is:
1) v/8a
2) 2a✓4+π^2
3) a
4) a/2✓4+π^2
Answers
Answer:
tangential acceleration is asked... It is = the derivative of the tangential velocity.
acceleration = dv / dt = d (a t ) / dt = a
this is always constant.
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Completion of 1/8 of revolution means: Ф = angle turned = π/4 radians, or the distance traveled is = 2πR / 8 = π R/4
t = time
v = instantaneous linear speed along the circular arc = a * t
radius = R
ω = instantaneous angular velocity = v / r = a t / R
linear displacement s = integral v * dt
So s = a t² / 2, we choose s = 0 at t = 0
= π R / 4
=> t² = (πR) / (2a)
t = √ [ πR / (2a) ]
This is the time taken to reach the point where the body covered 1/8 of the revolution.
tangential or Linear acceleration = dv/dt = a, it is always a constant.
centripetal acceleration = v² / R = a² t² / R
resultant acceleration will be the net result of the two components.
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Explanation: