Use the given equation for magnitude of drag force to calculate time taken

In summary, the problem involves a ball of mass 0.5 kg experiencing a drag force of DT(t) = 0.01[v(t)]^2 while moving horizontally with a speed of 10 m/s, which reduces to 8 m/s over a time interval t. The drag force is determined by the velocity squared and always opposes the direction of motion. Using the equation D = bv^2, where b is a constant, and v(t) = √(mg/b), an estimated answer of 23 m/s after 3 seconds is obtained, but it may not be accurate due to the changing drag force as the velocity decreases.
  • #1
bobpeg123
8
0

Homework Statement



A ball of mass 0.5 kg traveling horizontally through the air with speed v(t) experiences
a drag force DT (t) whose magnitude is given by:
DT (t ) = 0.01[v (t )] 2
The speed of the ball is found to reduce from 10 m/s to 8 m/s over a time interval t,
during which time the ball continues to travel horizontally. Calculate t.


Homework Equations


I'm not entirely sure where to start with this. My notes give me the equation for drag forces falling vertically but I wasn't sure what to do because it's traveling horizontally.


The Attempt at a Solution


 
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  • #2
bobpeg123 said:

Homework Statement



A ball of mass 0.5 kg traveling horizontally through the air with speed v(t) experiences
a drag force DT (t) whose magnitude is given by:
DT (t ) = 0.01[v (t )] 2
The speed of the ball is found to reduce from 10 m/s to 8 m/s over a time interval t,
during which time the ball continues to travel horizontally. Calculate t.


Homework Equations


I'm not entirely sure where to start with this. My notes give me the equation for drag forces falling vertically but I wasn't sure what to do because it's traveling horizontally.
Why would it be different for something moving horizontally?

Remember, the drag force always points in the opposite direction of the velocity.
 
  • #3
But does that mean that the mass isn't important, or that gravity therefore doesn't matter?
 
  • #4
Not for drag force. For a given object in a given fluid (air), the drag force is completely determined by only two things: how fast the object is moving (drag force is proportional to velocity squared) and which way it's moving (drag force always opposes the velocity). It doesn't matter whether gravity is in effect or not.
 
  • #5
Ok, so I tried again and I got,

for v = 100m/s, D = 10N,

using D = bv^2, b=10/100^2 = 0.001

and then using v(t) = {sqrt(mg/b)... expotential stuff, which gave me an answer of 23m/s after 3 secs, but I don't know if that seems reasonable because as the velocity drops, so does the drag so how would it drop 77m/s in 3 seconds.
 

1. How is the magnitude of drag force calculated?

The magnitude of drag force is calculated using the equation Fd = (1/2) * ρ * v^2 * Cd * A, where ρ is the density of the fluid, v is the velocity of the object, Cd is the drag coefficient, and A is the cross-sectional area of the object.

2. What is the significance of the given equation in calculating the time taken?

The given equation for magnitude of drag force is used to calculate the force acting on an object due to air resistance, which is known as drag. This force directly affects the object's velocity, thus making it an important factor in determining the time taken for the object to reach a certain distance.

3. Can the given equation be used for all types of objects?

Yes, the given equation can be used for any object moving through a fluid, whether it is a solid object like a car or a fluid object like a drop of liquid. However, the values of ρ, v, Cd, and A may vary depending on the object's shape and the characteristics of the fluid it is moving through.

4. How does changing the variables in the equation affect the time taken?

Changing the variables in the equation will directly affect the magnitude of drag force, which in turn will affect the object's velocity. A higher drag force will result in a slower velocity, thus increasing the time taken for the object to travel a certain distance. Similarly, a lower drag force will result in a higher velocity and a shorter time taken.

5. Is the given equation accurate for all situations?

The given equation is an idealized model and may not be completely accurate in all situations. Factors such as turbulence, air pressure, and temperature can affect the magnitude of drag force and the object's velocity, thus potentially affecting the calculated time taken. However, the equation provides a good estimate for most practical situations.

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