+117 I'm having trouble understanding how to solve this equation. Could anyone please help me solve this problem?
A little ball made of steel is dropped from a height. The speed of the ball kan be described with the formula v(t) = 0.028 m/s * (1-e ^ -300s-1*t)
a) Find the terminal speed
b) Find acceleration and the position of the ball (a(t) + s(t) ) I think
+33657 I'm going to assume you mean the velocity as a function of time is \(v(t)=0.028(1-e^{-300t})\)
where t is in seconds and velocity is in m/s.
a) Terminal velocity will be when t is infinite, so e^(-300t) → 0 and vterminal → 0.028 m/s
b) Acceleration: \(a(t)=\frac{dv(t)}{dt}\quad a(t)=300\times0.028e^{-300t}\rightarrow8.4e^{-300t}\) acceleration will be in m/s^2
Position: \(s(t)=\int v(t)dt\quad s(t)=0.028(t+\frac{e^{-300t}}{300})+s(0)\) where s(0) is position at time 0.
+33657 I'm going to assume you mean the velocity as a function of time is \(v(t)=0.028(1-e^{-300t})\)
where t is in seconds and velocity is in m/s.
a) Terminal velocity will be when t is infinite, so e^(-300t) → 0 and vterminal → 0.028 m/s
b) Acceleration: \(a(t)=\frac{dv(t)}{dt}\quad a(t)=300\times0.028e^{-300t}\rightarrow8.4e^{-300t}\) acceleration will be in m/s^2
Position: \(s(t)=\int v(t)dt\quad s(t)=0.028(t+\frac{e^{-300t}}{300})+s(0)\) where s(0) is position at time 0.
