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
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.
Well, let me start by saying that I'm not really the best when it comes to physics problems, but I'll give it a shot and try to keep it as light-hearted as possible!
To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force acting on the motorboat is the force of the drag caused by the water.
The drag force is proportional to the velocity, so we can express it as F = -kv, where k represents the constant of proportionality.
Now, let's break down the problem step by step:
1. At t = 0, the motorboat shuts off its engines, so there is no external force acting on it. This means that the net force on the motorboat is zero.
2. The net force on the motorboat is equal to the drag force, so we can write -kv = 0. Since we're interested in finding the distance traveled, we'll need to solve for v first.
3. Rearranging the equation, we get: v = 0.
Now, this might be a bit anticlimactic, but since the boat comes to rest in 2.7 seconds, the distance traveled can be calculated by simply multiplying the average velocity by the time. In this case, the average velocity is half of the initial velocity, which is 2.1 m/s (as you mentioned). So, the distance traveled would be:
distance = average velocity * time = (2.1 m/s / 2) * 2.7 s = 2.835 meters.
And that's it! The motorboat travels approximately 2.835 meters before coming to rest.
I hope this helps, but please keep in mind that my calculations might not be 100% accurate. It's always a good idea to double-check the work or ask for help from a physics expert. Stay afloat, my friend!
To find the distance traveled, we need to integrate the velocity function over time.
Given that the drag force is proportional to velocity (v), we can use the following differential equation:
m * dv/dt = -k * v
Where:
m = mass of the motorboat (assumed constant)
k = proportionality constant (related to the drag force)
v = velocity of the motorboat at any given time t
Rearrange the equation to separate variables:
dv/v = (-k/m) * dt
Integrate both sides:
∫(1/v) dv = ∫(-k/m) dt
ln|v| = (-k/m) * t + C
Apply the initial condition v(0) = 2.1 m/s at t = 0:
ln|2.1| = (-k/m) * 0 + C
ln|2.1| = C
The velocity function becomes:
ln|v| = (-k/m) * t + ln|2.1|
Taking the exponential of both sides to eliminate the natural logarithm:
|v| = e^((-k/m) * t + ln|2.1|)
Since speed (v) cannot be negative, we can drop the absolute value:
v = e^((-k/m) * t + ln|2.1|)
Now, we are given that after 2.7 seconds (t = 2.7 s), the speed has dropped to half its original value.
Substitute t = 2.7 s and v = 2.1 m/s into the velocity function:
2.1 = e^((-k/m) * 2.7 + ln|2.1|)
Solve this equation for the ratio (-k/m):
(-k/m) = ln(2.1/2.1e^2.7)
Simplify:
(-k/m) = -2.7
Now, we have the velocity function:
v = e^(-2.7t + ln|2.1|)
To find the time it takes for the motorboat to come to rest (v = 0), set the velocity function equal to zero:
0 = e^(-2.7t + ln|2.1|)
Take the natural logarithm of both sides:
ln(0) = -2.7t + ln|2.1|
Since the natural logarithm of zero is undefined, it means the motorboat will never come to rest. However, if we consider the limiting case as time approaches infinity, the velocity will approach zero.
To find the distance traveled before coming to rest, we need to integrate the velocity function over time:
∫v dt = ∫e^(-2.7t + ln|2.1|) dt
Let u = -2.7t + ln|2.1|:
du = -2.7 dt
Substituting back into the integral:
(1/(-2.7)) ∫e^u du = (1/(-2.7)) e^u + C
Substituting back for u:
(1/(-2.7)) e^(-2.7t + ln|2.1|) + C
Since the motorboat starts at t = 0 with speed 2.1 m/s, the constant C can be determined by substituting these values:
2.1 = (1/(-2.7)) e^(-2.7 * 0 + ln|2.1|) + C
Solve this equation for C:
C = 2.1 - (1/(-2.7)) e^ln|2.1|
Simplify:
C = 2.1 + (1/(-2.7)) * 2.1 = 2.1 - 0.7778 ≈ 1.322
The distance traveled before coming to rest is given by the definite integral:
∫[0 to ∞] v dt = [(1/(-2.7)) e^(-2.7t + ln|2.1|)] [0 to ∞]
Substituting the limits:
[1/(-2.7)] e^(∞ + ln|2.1|) - [1/(-2.7)] e^(0 + ln|2.1|)
As time (t) approaches infinity, e^(-2.7t) approaches zero. Therefore, the first term becomes zero:
0 - [1/(-2.7)] e^ln|2.1|
Simplifying:
(1/2.7) * 2.1 ≈ 0.7778
Therefore, the motorboat travels approximately 0.7778 meters before coming to rest.
To solve this problem, we can use the laws of motion and the concept of exponential decay. Let's break down the steps:
1. Define the variables:
- Let v0 be the initial velocity of the boat (2.1 m/s).
- Let v(t) be the velocity of the boat at time t.
- Let k be the proportionality constant for the drag force.
2. Write down the differential equation:
In this case, the drag force is proportional to the velocity of the boat. We can use Newton's second law to write the differential equation:
m * dv/dt = -k * v
3. Solve the differential equation:
Rearranging the equation:
dv/v = -(k/m) dt
Integrating both sides:
∫(dv/v) = -k/m ∫dt
Integrating from v0 to v(t) and from 0 to t:
ln(v(t)) - ln(v0) = -k/m * t
Using the property ln(a) - ln(b) = ln(a/b):
ln(v(t)/v0) = -k/m * t
Exponentiating both sides:
v(t)/v0 = e^(-k/m * t)
Solving for v(t):
v(t) = v0 * e^(-k/m * t)
4. Find the value of k/m:
We are given that the boat's speed has dropped to half its original value after 2.7 seconds. Thus, v(2.7) = v0/2.
Plugging in the values:
v(2.7) = v0 * e^(-k/m * 2.7)
v0/2 = v0 * e^(-k/m * 2.7)
Divide both sides by v0:
1/2 = e^(-k/m * 2.7)
Apply the natural logarithm to both sides:
ln(1/2) = -k/m * 2.7
Solve for k/m:
-2.7 * (k/m) = ln(1/2)
k/m = -ln(1/2) / 2.7 ≈ 0.405
5. Substitute k/m and v0 into the equation:
v(t) = v0 * e^(-k/m * t)
v(t) = 2.1 * e^(-0.405 * t)
6. Find the time when the velocity of the boat is zero:
When the boat comes to a stop, v(t) = 0. We can solve this equation to find the time it takes for the boat to stop:
0 = 2.1 * e^(-0.405 * t)
e^(-0.405 * t) = 0
Take the natural logarithm of both sides:
-0.405 * t = ln(0)
However, it is not possible to take the natural logarithm of 0, which means the boat will never come to a complete stop in this scenario.
Therefore, we cannot determine the distance the boat will travel before coming to rest using the given information.
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.