Question

use tangential and normal acceleration to find the magnitude

Problem 7: A vehicle travels along a circular track as shown in the diagram with a constant speed of 14 m/s. Just at the point when the vehicle reaches point A it increases its speed according to the following expression, at Determine the magnitudes of the velocity and acceleration of the vehicle when it reaches point B Also, how much time will elapse as it travels from A to B? -(v/3 m/s, where v is in m/s. 4 200 m

Answer 1

The distance travelled by the car during its journey from A to B. s = 2 pi R/4 = pi(200 m)/2 = 314 The acceleration is given as follows: a_t = dv/dt = (ds/dt)(dv/ds) = v dv/ds Therefore, v dv/ds = (5/4 v^1/3) v dv/v^1/3 = 5/4 ds Integrating both sides, integral^v_b_14 v^2/3 dv = integral^314_0 (5/4) ds [3/5 v^5/3]^v_14 = [5/4 s]^314_0 3/5 v^5/3 - 3/5 (14)^5/3 = 5/4(314) = 392.5 Rearrange the above-mentioned equation for v. v = ((392.5(5/3) + 3/2 (14)^5/3))^3/5 = 52.49 m/s Use dv/dt = 5/4 v^1/3 to find time taken to reach the point B by the car as follows: dv = 5/4 v^1/3 dt Integrating both sides. Integral^52.49_14 dv/v^1/3 = integral^t_0 5/4 dt 3/2 [v^2/3]^52.49_14 = 5/4 t 12.31 = 5/4 t Therefore, t = 9.85 s The tangential acceleration at point B is given as follows: a_t = (5/4 v^1/3) = (5/4 (52.49)^1/3) = 4.68 m/s^2 The centripetal acceleration is