Question

3. A particle moves in a circle of radius 1'0 cm at a speed
given by v = 2:0 t where v is in cm/s and t in seconds.
(a) Find the radial acceleration of the particle at t = ls.
(b) Find the tangential acceleration at t = 1 s. (c) Find
the magnitude of the acceleration at t = 1 s.
11​

Answer 1

Answer:

Explanation:

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Answer 2

Given :-

Radius = 1.0 cm

Velocity = 2t

To Find :-

The radial acceleration of the particle at t = 1 s

The tangential acceleration at t = 1 s

The magnitude of the acceleration at t = 1 s

Solution :-

We know that,

• u = Initial velocity
• r = Radius
• dv/dt = Volts per second increase or decrease

Given that,

Radius = 1.0 cm

Velocity = 2t

Solution I :

At t = 1s

Velocity of the particle, u

$\sf 2.0 \: t = 2.0 \times 1 = 2.0 \: cm/s$

Radial acceleration of the particle at t

$\sf 1 \ is \ \dfrac{v^{2}}{2}$

Substituting them, we get

$\sf \dfrac{(2.0)^{2}}{1}=4 \ cm/s^{2}$

Therefore, the radial acceleration of the particle at t = 1 s is 4 cm/s²

Solution II :

We know that,

$\sf a=\dfrac{dv}{dt} =\dfrac{d}{dt} (2t)= 2 \ cm/s^{2}$

Tangential acceleration at t

$\sf 1 \ is \ 2 \ cm/s^{2}$

Therefore, the tangential acceleration of the particle at t = 1 s is 2 cm/s²

Solution III :

Magnitude of acceleration at t = 1 sec

$\sf Magnitude \ of \ acceleration=\sqrt{(Radial \ acceleration)^{2}+(Tangential \ acceleration)^{2}}$

Substituting their values, we get

$\sf Magnitude \ of \ acceleration=\sqrt{4^{2}+2^{2}}$

$\sf =\sqrt{20} \ cm/s^{2}$

Therefore, the magnitude of the acceleration at t = 1 s is √20 cm/s²