# Solve Drag Force Problem: Find Time to Reach 2% of Orig. Speed

• Symstar
In summary, Homework Equations state that ma=F_D-mg, where F_D is the drag force and m is the mass. The equation also states that v(t)=v_0+at, where v_0 is the initial velocity and at is the time at which the drag force is applied. When v decreases, the acceleration does as well, which creates a problem when dealing with non-constant accelerations. To solve this problem, you need to use partial derivatives. Additionally, when determining the time until 2% of the initial velocity is reached, you need to use the equation v(t)=v_0+at and solve for t.
Symstar

## Homework Statement

You dive straight down into a pool of water. You hit the water with a speed of 6.5 m/s, and your mass is 65 kg.

Assuming a drag force of the form $$F_D=cv=(-1.2*10^4 \tfrac{kg}{s})*v$$, how long does it take you to reach 2% of your original speed? (Ignore any effects of buoyancy.)

## Homework Equations

$$ma=F_D-mg$$
$$v(t)=v_0+at$$

## The Attempt at a Solution

$$ma=F_D-mg$$
$$a=\frac{F_D}{m}-g$$
$$a=\frac{(-1.2*10^4)*-6.5}{65}-9.8=1190.2$$

$$v(t)=v_0+at$$
$$-6.5*.02=-6.5+1190.2t$$
$$t=5.4*10^{-3}$$

Logically, the answer does not make sense, nor is it correct. Where did I go wrong?

Last edited:
You have assumed acceleration is constant. In your equation, $$a = \frac{cv}{m} - g$$

And then you used constant acceleration formulae. However, think about what happens as v changes.

Rake-MC said:
You have assumed acceleration is constant. In your equation, $$a = \frac{cv}{m} - g$$

And then you used constant acceleration formulae. However, think about what happens as v changes.

Ah, yes. As velocity decreases, the acceleration does as well. This is problematic for me, however, as I have not worked with non-constant accelerations. How should I go about solving this problem?

Try this:
solve for Fnet first, before acceleration.

If you mean:
$$F_{net}=cv-mg$$

I can solve for it, but do I use 6.5 as my value for v? I still don't know what to do with the resulting value? Doesn't that still have the same problem of assuming constant acceleration?

$$F = m \frac{v}{t}$$so $$m \frac{v_s}{t} = cv - mg$$

where v_s is the one which is at 2%
re arrange to get t.

I hope I didn't make any careless mistakes I was up all night

$$F_{net}=cv-mg=(-1.2*10^4)(-6.5)-65(9.8)=77363$$

$$F_{net}=m\tfrac{v_s}{t}$$
$$77363=65(\tfrac{0.13}{t}$$
$$t=1.09*10^{-4}$$

This is also an incorrect answer =/

Shameless bump - still haven't resolved this question :(

You have to use partial derivatives.

FD-mg = ma
cv-mg = m dv/dt
cv/m-g=dv/dt
integral(dt)|0,t = (dv/ (cv/m)-g)

Ahh catamara is totally right, we were on the right track though

## 1. What is drag force and how does it affect an object's speed?

Drag force is a type of resistance force that acts opposite to an object's motion through a fluid, such as air or water. This force can slow down the object's speed as it moves through the fluid.

## 2. How do you calculate drag force?

Drag force can be calculated using the formula Fd = 0.5 * 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.

## 3. What is the significance of finding the time to reach 2% of the original speed?

This calculation can be useful in determining the impact of drag force on an object's speed over time. It can also be used to compare the effects of different fluids or objects on drag force.

## 4. How do you find the time to reach 2% of the original speed?

The time to reach 2% of the original speed can be found by using the equation t = (2 * m) / (ρ * v * A * Cd), where m is the mass of the object. This equation can be derived from the drag force formula by setting Fd equal to 2% of the original speed.

## 5. What factors can affect the time it takes to reach 2% of the original speed?

The time to reach 2% of the original speed can be affected by factors such as the object's mass, the density of the fluid it is moving through, its initial velocity, the drag coefficient of the object, and the object's cross-sectional area. Additionally, external factors such as wind or changes in the fluid's density can also impact this time calculation.

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