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
What acceleration does the parachutist experience just after opening the parachute? (m/s^2)
v = 18 - 142 e^-(t/2.5)
a = dv/dt = (142/2.5) e^-(t/2.5)
when t = 0
a = 142/2.5
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
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.
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.