Explanation:
The position of the car at any instant is defined from the fixed point. The velocity and acceleration are to be determined at a point B (i.e s=3) . So, the origin of the n,t axes at this point (Check diagram)
Velocity:
The bicyclist travels around a horizontal circular path at a speed of v = (0.09t2 + 0.1t) m/s then the rate of increase in speed is
v= (0.18t-F 0.1) m/s2 (1)
The time needed for the bicycle to reach point B can be determined by realising that the position of B is sb = 3 m and since sA = 0 when t = 0 we have
v = ds/dt= (0.09t2 + 0.1t)) m/s (2)
[tex]\int\limits^3_0 { \, ds[/tex] = [tex]\int\limits^t_0 {(0.09t2 + 0.1t)} \, dt[/tex]
3 m= 0.03[tex]t^{3}[/tex]B + 0.05tB
tB= 4.15 s
Substituting into Eq. 2 yields
vg = (0.09(4.15)2+0.1(4.15)) m/s
= 1.965 m/s
Acceleration:
The tangential component of acceleration at B is obtained by substituting tb= 4.15 s into Eq. 1
(ab)n = [tex]v^{2}[/tex]B / p = (0.18(4.15+ 0.1 )m/s² = 0.847 m/s²
Realising that the circular path is of radius p=10 m and using vb=1.985 m/s the normal component of acceleration at B is therefore
(ab)n = [tex]v^{2}[/tex]B / p = (1.965 m/s )² / 10m = 0.386 m/s²
The magnitude of acceleration is therefore given by
ab = [tex]\sqrt{(0.847 m/s ^{2} )^{2} + (0.386 m/^{2} s)^{2}[/tex] => [tex]a_{b}[/tex] = 0.93 [tex]ms^{2}[/tex]
