nonuniform circular motion?

Question

a particle is moving on a circular path such that tangential acceleration is equal to the normal acceleration. at time t=0 velocity is u find the velocity after time t

Answer 1

Given
a (tangential) = a (radial) = v^2 / r
{Magnitudes alone can be equal}

dv/dt = v^2 / r

dv / v^2 = dt / r

Integrating both sides

- 1/v = t /r + C

Let at time t = 0, v = u. C = - 1/u

- 1/v = t /r - 1/u

- 1/v = t /r- 1/u

v = ur / [r –ut]

Answer 2

The assumption that the tangential acceleration is equal to the radial acceleration just doesnt exist in orbital motion.
An example is as follows.
The Earth which orbits the Sun in a quasi Circular orbit has a much greater tangential aceleration than radial acceleration relative to the Sun.
If this was the question in an exam it probably meant Velocity rather than acceleration.

Acceleration is always maximum at the begining of a body 's motion.At time zer0 the velocity is Zero . At time zero plus, velocity is born and increases as acceleration attenuates with time .Till a final velocity is reached.

Answer 3

Here's a hint. If the tangential acceleration and the normal acceleration are equal, then that means the change in velocity in those directions is also equal. That could only be the case for uniformly accelerating motion. To see this draw a vector triangle and see that it is a 45-45-90 degree triangle. If you know the motion along the circular arc is uniformly accelerating, use the equations you studied earlier for uniformly accelerating motion along a line to solve for s.

Answer 4

if it is around action then despite that's is going around in a circle? so the sole non-uniformity may be the angular velocity? for this reason centripetal acceleration is v^2/r in the two case. yet possibly you advise some thing else, like non-around.

Answer 5

object covering circular path in unequal interval of time