A falling object satisfies the initial value problem dv/dt=9.8-(v/5), v(0)=0
wherevis the velocity in meters per second.
(a)Find the time, in seconds, that must elapse for the object to reach 97% of its limiting velocity.
Round your answer to two decimal places.
t equalss
(b)How far, in meters, does the object fall in that time?
Round your answer to two decimal places.
The Correct Answer and Explanation is :
Let’s go through the solution for each part step-by-step:
Given:
The differential equation is:
[
\frac{dv}{dt} = 9.8 – \frac{v}{5}
]
where ( v(t) ) is the velocity of the object at time ( t ), and the initial condition is ( v(0) = 0 ).
(a) Find the time it takes for the object to reach 97% of its limiting velocity.
The limiting velocity (also called terminal velocity) occurs when the acceleration (( \frac{dv}{dt} )) becomes zero. Setting ( \frac{dv}{dt} = 0 ) in the differential equation, we get:
[
0 = 9.8 – \frac{v}{5}
]
Solving for ( v ), we get:
[
v = 9.8 \times 5 = 49 \, \text{m/s}
]
So the limiting velocity is 49 m/s.
Now, we need to find the time when the velocity reaches 97% of 49 m/s:
[
0.97 \times 49 = 47.53 \, \text{m/s}
]
Solution of the differential equation:
Rewriting the differential equation:
[
\frac{dv}{dt} = 9.8 – \frac{v}{5}
]
To solve this, we separate the variables:
[
\frac{dv}{9.8 – \frac{v}{5}} = dt
]
Now, integrate both sides. First, rewrite the left side to make it easier to integrate:
[
\frac{dv}{9.8 – \frac{v}{5}} = \frac{5}{49 – v} \, dv
]
This can be integrated using the substitution ( u = 49 – v ), which gives us:
[
\int \frac{5}{u} \, du = \int dt
]
After integrating:
[
5 \ln |49 – v| = t + C
]
Using the initial condition ( v(0) = 0 ), we can solve for ( C ):
[
5 \ln 49 = 0 + C \quad \Rightarrow \quad C = 5 \ln 49
]
So, the solution for ( v(t) ) is:
[
v(t) = 49 – 49 e^{-t/5}
]
Time for 97% of limiting velocity:
We want ( v(t) = 47.53 ):
[
47.53 = 49 – 49 e^{-t/5}
]
[
e^{-t/5} = \frac{49 – 47.53}{49} = 0.03
]
Taking the natural logarithm of both sides:
[
-\frac{t}{5} = \ln 0.03
]
[
t = -5 \ln 0.03 \approx 5 \times 3.5065 = 17.53 \, \text{seconds}
]
Thus, the time required to reach 97% of the limiting velocity is approximately 17.53 seconds.
(b) How far does the object fall in that time?
To find the distance, we use the relationship between velocity and distance. The velocity is:
[
v(t) = 49 – 49 e^{-t/5}
]
To find the distance traveled, we integrate the velocity:
[
s(t) = \int_0^t v(t) \, dt = \int_0^t \left( 49 – 49 e^{-t/5} \right) dt
]
This can be split into two integrals:
[
s(t) = \int_0^t 49 \, dt – \int_0^t 49 e^{-t/5} \, dt
]
The first integral is straightforward:
[
49t
]
The second integral is:
[
49 \times 5 \left( -e^{-t/5} \right) = -245 e^{-t/5}
]
So the total distance is:
[
s(t) = 49t + 245 \left( 1 – e^{-t/5} \right)
]
Substitute ( t = 17.53 ) into the equation:
[
s(17.53) = 49(17.53) + 245 \left( 1 – e^{-17.53/5} \right)
]
Now, compute the result:
[
s(17.53) \approx 49 \times 17.53 + 245 \times (1 – 0.03)
]
[
s(17.53) \approx 858.97 + 245 \times 0.97
]
[
s(17.53) \approx 858.97 + 237.65 \approx 1096.62 \, \text{meters}
]
Thus, the object falls approximately 1096.62 meters in the time it takes to reach 97% of its limiting velocity.
Final Answers:
- (a) The time is 17.53 seconds.
- (b) The distance fallen is 1096.62 meters.