A particle moves along a curved path. At O the speed is 12 ft/s and at A the spe

Question

A particle moves along a curved path. At O the speed is 12 ft/s and at A the speed has slowed to 6ft/s. The distance OA along the path is 18ft. If the particle has deceleration proportional to the distance along the path from O, and the magnitude of total deceleration at A of 10ft/s, determine the deceleration value at A and the radius of curvature () of the path at A.

I have the solution to this problem but im lost on what formulas they're using. Can you please do this problem step by because i want to know exactly how to do this problem and exactly what formulas are being used. I will give you a lifesaver if you can do this. Thank You!!

Explanation / Answer

Hmm... I will give it a shot.

I am assuming that by "magnitude of total deceleration at A of 10ft/s" you really mean "10 ft/s^2" ? If so, then here is my answer...

The deceleration is proportional to distance traveled, so the acceleration (t axis) is negative and constant. Solve for dv/ds

dv/ds = (change in velocity)/(change in position) = (6-12)/(18) = -1/3

at point A, a_t = v(dv/ds) = 6(-1/3) = -2

a = sqrt ( a_t^2 + a_n^2), so

10 = sqrt [ (-2)^2 + a_n^2 ]

100 = 4 + a_n^2

a_n = sqrt (96) 9.80 ft/s^2

a_n = (v^2)/rho => rho = (v^2)/a_n

rho = (6^2)/9.80 = 36/9.80

rho = 3.67 ft.

Am i close? The question is certainly worded in a confusing fashion...

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.