At some instant, a particle traveling in a horizontal circular path of radius 5.90 m has a total acceleration with a magnitude of 18.0 m/s2 and a constant tangential acceleration of 12.0 m/s2. determine the speed of the particle at this instant and (1/8) revolution later.

The final speed will be 8.02 rad/s.

Explanation:

Given that

radius ,r=5.9 m

Total acceleration = 18

Tangential acceleration = 12

Here particle is moving in circular path with varying linear speed.So there will be two acceleration ,one is radial acceleration and other one is tangential acceleration.

12=α x 5.9

ω=2.2 rad/s

θ=2πn

θ=2 x π x 1/8 rad

θ=π /4 rad

So the final speed will be 8.02 rad/s.

v 154.9= 0.9 length length s little over 154.9cm

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west