TIME SENSITIVE! Dynamics problem!?
A particle travels around a circular path having a radius of 64m. It is initially traveling with a speed of 10m/s and its speed then increases at a rate of v[dot] = (0.05v) m/s^2.
Determine the magnitude of the particle's acceleration four seconds later.
Comments
v̇ = dv/dt = 0.05 × v
Therefore dv/v= 0.05 × dt. Integrating both sides:
ln(v) = 0.05 × t + c'
v = C.℮^(0.05 × t)
Now at t=0, v = 10 m/s therefore C=10 and at t=4
v = 10 × ℮^(0.05 × 4) = 10×℮^(0.2) = 12.214 m/s
The particle's linear acceleration is then 0.05 × v = 0.05 × 12.214 = 0.6107 m/s².
The particle's radial acceleration = v²/r = (12.214)²/64 = 2.331 m/s²
Giving a total acceleration = √(0.6107² + 2.331²) = 2.41m/s²