The velocity of a parachutist as a function of time is given by v = vf + (v0 - vf)e-t/2.5, where t = 0 s corresponds to the instant the parachute is opened, v0 = 160 km/h is the velocity before opening of the parachute, and vf = 18 km/h is the final (terminal) velocity.
What acceleration does the parachutist experience just after opening the parachute? (m/s^2)
whoops,need to change km/hour to meters/second
meters/second = km/hour * 1000 m/km * 1 hr/3600 seconds
= km/hr / 3.6
vo = 160/3.6 = 44.4 m/s
vf = 18/3.6 = 5 /s
then do it
To find the acceleration experienced by the parachutist just after opening the parachute, we need to take the derivative of the velocity equation with respect to time (t).
Given:
v = vf + (v0 - vf)e^(-t/2.5)
v0 = 160 km/h = 160 * 1000 m / 3600 s = 44.44 m/s
vf = 18 km/h = 18 * 1000 m / 3600 s = 5 m/s
To find the derivative, we have:
dv/dt = d/dt [vf + (v0 - vf)e^(-t/2.5)]
= 0 + d/dt [(v0 - vf)e^(-t/2.5)]
= (v0 - vf) * d/dt [e^(-t/2.5)]
= (v0 - vf) * (-1/2.5) * e^(-t/2.5)
Plugging in the given values:
dv/dt = (44.44 m/s - 5 m/s) * (-1/2.5) * e^(-t/2.5)
Now, to find the acceleration, we can substitute the value of t = 0, as we want to find the acceleration just after opening the parachute:
a = (44.44 m/s - 5 m/s) * (-1/2.5) * e^(-0/2.5)
= 39.44 m/s * (-1/2.5) * e^0
= -15.78 m/s^2
Therefore, the parachutist experiences an acceleration of approximately -15.78 m/s^2 just after opening the parachute.
To find the acceleration experienced by the parachutist just after opening the parachute, we need to differentiate the given velocity function with respect to time.
Given: v = vf + (v0 - vf)e^(-t/2.5)
To find the acceleration, we differentiate the velocity function v with respect to time t:
a = dv/dt
Let's differentiate the function:
a = d/dt [vf + (v0 - vf)e^(-t/2.5)]
The constant term, vf, differentiates to zero since it's a constant.
The differentiation of the second term, (v0 - vf)e^(-t/2.5), will require the chain rule.
Let's differentiate the second term:
a = d/dt [(v0 - vf)e^(-t/2.5)]
= (v0 - vf)(d/dt [e^(-t/2.5)])
The derivative of e^(-t/2.5) can be found by applying the chain rule.
Let u = -t/2.5, then du/dt = -1/2.5.
Using the chain rule, we have:
d/dt [e^(-t/2.5)] = (du/dt) * (d/du [e^u])
= (-1/2.5) * e^u
= (-1/2.5) * e^(-t/2.5)
Substituting this result back into our equation for a:
a = (v0 - vf)(d/dt [e^(-t/2.5)])
= (v0 - vf)(-1/2.5) * e^(-t/2.5)
Now, we can simplify further:
a = (v0 - vf)(-1/2.5) * e^(-t/2.5)
= (v0 - vf) * (-2/5) * e^(-t/2.5)
Since the question asks for the acceleration in units of m/s^2, we need to convert the velocities, v0 and vf, from km/h to m/s before calculating the acceleration.
v0 = 160 km/h * (1000 m/km) * (1 h/3600 s)
= 44.4444... m/s
vf = 18 km/h * (1000 m/km) * (1 h/3600 s)
= 5 m/s
Now we can substitute the given values into the equation for acceleration:
a = (44.4444... - 5) * (-2/5) * e^(-t/2.5)
And this is the equation for the acceleration experienced by the parachutist just after opening the parachute, in m/s^2.
v = 18 + (160 - 18)e^-(t/2.5)
v = 18 - 142 e^-(t/2.5)
a = dv/dt = (142/2.5) e^-(t/2.5)
when t = 0
a = 142/2.5