Answer to Question #177979 in Differential Equations for Umer aziz

Question

A bike is accelerating in yz-plane with its speed given by ()

at

t(w2) +2 * u2: dvt)

at

= u2

= u3 * t(2+w3)+3*u3, subject to the initial conditions,

v,(0) = n2; v0) = n3. Determine its speed at later time

u2 = 4 + 0.3R, u3 = 2+ 0.4R, w2 = 0.5+ R,w3 = 1+ R, n2 = 3.1+ 0.2R, n3 = 4.1 + 0.1R, p1 = 5.4 + 0.2R

Answer 1

$\frac{\partial v_y(t)}{\partial t}=u_2\cdot3t^{w_2},\ \frac{\partial v_z(t)}{\partial t}=u_3\cdot t^{w_3}+3$

$v_y(0)=n_2,\ v_z(0)=n_3$

$v_y(t)=\frac{3u_2t^{w_2+1}}{w_2+1}+n_2$ , $v_z(t)=\frac{u_3t^{w_3+1}}{w_3+1}+n_3$

$v_y(p_1)=\frac{3(3+0.3R)(5.4+0.2R)^{2+0.3R}}{2+0.3R}+3.1+0.2R$

$v_z(p_1)=\frac{(5+0.4R)(5.4+0.2R)^{2+R}}{2+R}+4.1+0.1R$