+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)=dv(t)dta(t)=300×0.028e−300t→8.4e−300t acceleration will be in m/s^2
Position: s(t)=∫v(t)dts(t)=0.028(t+e−300t300)+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)=dv(t)dta(t)=300×0.028e−300t→8.4e−300t acceleration will be in m/s^2
Position: s(t)=∫v(t)dts(t)=0.028(t+e−300t300)+s(0) where s(0) is position at time 0.
