Starting from rest, a bicyclist travels around a horizontal circular path, r = 10 m, at a speed of v = (0.09t2 + 0.1t) m>s, where t is in seconds. Determine the magnitudes of his velocity and acceleration when he has traveled s = 3 m.

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)  

=

3 m= 0.03B + 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  = 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 = B / p = (1.965 m/s )² / 10m = 0.386 m/s²

The magnitude of acceleration is therefore given by

ab = => = 0.93

do you have a formula possibly