A particle moves in a circle of radius 2.0 m at a speed given by v = 2t, where v is in ms–1 and t is in second. Find the magnitude of acceleration of particle at t = 1 s.Only one correct answerA.
Question
A particle moves in a circle of radius 2.0 m at a speed given by v = 2t, where v is in ms–1 and t is in second. Find the magnitude of acceleration of particle at t = 1 s.Only one correct answerA.
Solution
The acceleration of a particle moving in a circle is given by the sum of the radial acceleration (ar) and the tangential acceleration (at).
The radial acceleration is given by the formula ar = v^2/r, where v is the velocity and r is the radius of the circle.
The tangential acceleration is given by the formula at = dv/dt, where dv/dt is the rate of change of velocity.
Given that v = 2t and r = 2.0 m, we can substitute these values into the formulas to find the accelerations at t = 1 s.
First, find the velocity at t = 1 s: v = 2*1 = 2 m/s.
Then, find the radial acceleration: ar = v^2/r = (2 m/s)^2 / 2.0 m = 2 m/s^2.
Next, find the tangential acceleration: at = dv/dt = d(2t)/dt = 2 m/s^2.
Finally, find the total acceleration: a = sqrt(ar^2 + at^2) = sqrt((2 m/s^2)^2 + (2 m/s^2)^2) = sqrt(8) m/s^2 = 2.83 m/s^2.
So, the magnitude of the acceleration of the particle at t = 1 s is 2.83 m/s^2.
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