lets see if your model is correct:
v=vo*e^kt
dv/dt=a=kvo*e^kt
F=ma=m dv/dt=mkvo*e^kt=m*k*v
so it works, force is dependent on v.
Now solving:
at t=2.7, v=1/2 vo
1/2=e^2.7k
take the ln of feach side
-.693=2.7k
or k= = -0.257
distance=Integral v dt
=2.1 INT e^kt dt=2.1/k e^kt(from o to 2.7)
=2.1/k (1/2-1)= 4.09 meters.
check all that, I rounded some.
A motorboat traveling at a speed of 2.1 m/s shuts off its engines at t = 0. How far does it travel before coming to rest if it is noted that after 2.7 s its speed has dropped to half its original value? Assume that the drag force of the water is proportional to v.
Please show me the steps if you can. Right now I have the problem up to v=2.1e^3.896t but I'm not sure if it is right. Any help would be appreciated
2 answers
I know this is quite old but for any future souls seeking help, I struggled a lot on a similar problem but it's actually very simple.
For starters, your expression is correct.
v=vo*e^kt
Plugging in 2.1 m/s for vo, (2.1)/2 (or 1.05 m/s) for v, and 2.7 s for t, we can solve for k.
k=(ln(2.1/1.05))/(2.7) = -0.2567 1/s
Alternatively, because this time is the amount necessary for velocity to reduce to half its initial value, we can utilize the half-life expression, where k=-ln(2)/t
Because we want an expression for the distance traveled, we want to integrate our first velocity expression to find an expression for distance.
INT (vo*e^kt) = (vo/k)*e^kt
Because it took 2.7 seconds for the velocity to reduce to half of its original value, we can assume that this is the half-life at every moment. That is, it takes 2.7 seconds to reduce from 2.1 m/s to 1.05 m/s, 2.7 seconds to reduce from 1.05 m/s to 0.525 m/s, 2.7 seconds to reduce from .525 m/s to 0.2625 m/s, etc.
After about 10 half-lives, any system is going to achieve 99.9% of its goal. In this case, after 10 half-lives, the boat is going to cover 99.9% of the distance it is going to travel. It will technically keep moving forever, but for any practical calculation you can use this approach.
Using 10 half-lives, or 10*2.7 s, as our upper bound and 0 as our lower bound in the integral, 2.1 m/s for vo, and -0.2567 1/s for k, we get
distance = ((2.1/-0.2567)*e^(-0.2567)(27)) - ((2.1/-0.2567)*e^(-0.2567)(0))
=(-0.00799 m) - (-8.18076 m) = 8.17276 m
So the distance the boat travels before coming to rest, with two significant digits, is 8.2 meters. Hope this helps.
For starters, your expression is correct.
v=vo*e^kt
Plugging in 2.1 m/s for vo, (2.1)/2 (or 1.05 m/s) for v, and 2.7 s for t, we can solve for k.
k=(ln(2.1/1.05))/(2.7) = -0.2567 1/s
Alternatively, because this time is the amount necessary for velocity to reduce to half its initial value, we can utilize the half-life expression, where k=-ln(2)/t
Because we want an expression for the distance traveled, we want to integrate our first velocity expression to find an expression for distance.
INT (vo*e^kt) = (vo/k)*e^kt
Because it took 2.7 seconds for the velocity to reduce to half of its original value, we can assume that this is the half-life at every moment. That is, it takes 2.7 seconds to reduce from 2.1 m/s to 1.05 m/s, 2.7 seconds to reduce from 1.05 m/s to 0.525 m/s, 2.7 seconds to reduce from .525 m/s to 0.2625 m/s, etc.
After about 10 half-lives, any system is going to achieve 99.9% of its goal. In this case, after 10 half-lives, the boat is going to cover 99.9% of the distance it is going to travel. It will technically keep moving forever, but for any practical calculation you can use this approach.
Using 10 half-lives, or 10*2.7 s, as our upper bound and 0 as our lower bound in the integral, 2.1 m/s for vo, and -0.2567 1/s for k, we get
distance = ((2.1/-0.2567)*e^(-0.2567)(27)) - ((2.1/-0.2567)*e^(-0.2567)(0))
=(-0.00799 m) - (-8.18076 m) = 8.17276 m
So the distance the boat travels before coming to rest, with two significant digits, is 8.2 meters. Hope this helps.
1 answer