Answer to Question #260567 in Mechanics | Relativity for hamad

Question

5.2 A particle P moves along a parabolic path y= x, where x and y are expressed in meters. The particle

has a speed of 6 m/s which is increasing at 2 m/s' when x=6 m. Determine the direction of the velocity and

the magnitude and direction of the acceleration at this instant.

Answer 1

If the law is

$$y=\frac1{26}x^2,$$

at x=6 m we have

$$y=\frac1{26}6^2=0.72\text{ m}.$$

The direction of velocity is

$$\theta_v=\arctan\frac{y}{x}=6.86°$$

above +x axis.

Determine the radius of curvature at the point (6, 0.72), the centripetal acceleration, and the resultant acceleration:

$$R=\frac{[1+(y'(x))^2]^{3/2}}{|y''(x)|},\\\space\\ R=\frac{[1+(2/26·6)^2]^{3/2}}{2/26}=17.4\text{ m},\\ a_c=\frac{v^2}{R}=\frac{6^2}{17.4}=2.07\text{ m/s}^2,\\\space\\ a=\sqrt{a_c^2+a_\tau^2}=\sqrt{2.07^2+2^2}=2.88\text{ m/s}^2.$$