Question

The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the magnitude of the total acceleration is 6.0 m/s2, what is the speed of the particle? Group of answer choices

Answer 1

Answer:

The speed of the particle is 2.86 m/s

Explanation:

Given;

radius of the circular path, r = 2.0 m

tangential acceleration,  $a_t$ = 4.4 m/s²

total magnitude of the acceleration, a = 6.0 m/s²

Total acceleration is the vector sum of  tangential acceleration and radial acceleration

$a = \sqrt{a_c^2 + a_t^2}$

where;

$a_c$ is the radial acceleration

$a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2$

The radial acceleration relates to speed of particle in the following equations;

$a_c = \frac{v^2}{r}$

where;

v is the speed of the particle

$v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s$

Therefore, the speed of the particle is 2.86 m/s