Basic 1-d dynamics with a bit of calculus.

[sanhuy]

Homework Statement

A 500 kg boat moves with an initial speed of 20 m/s turns off it engine and slows down due to the drag force. If this force has magnitude F= 15.3v^2 find the boat speed after 12 seconds.

Homework Equations

∑F = ma
dv = adt

The Attempt at a Solution

taking the side opposite to the drag force to be positive in our force equation we get:
-15.3v² = m(-a) = m(-dv/dt).

so,
∫-15.3/500 dt = ∫ (1/v^2) dv with the correct limits i get V = -3.4m/s which is incorrect.

however, when i use the force equation
-15.3v² = m(a) = m(dv/dt) { no negative sign for dv/dt} i get the right answer v = 2.4 m/s .

I don't understand why this works. isn't acceleration in the same direction as the net drag force, so by how i set up my coordinate system they both should have a negative sign? Please help me understand. i looked online and in my textbook and i found nothing.

 

[ehild]

Homework Statement

A 500 kg boat moves with an initial speed of 20 m/s turns off it engine and slows down due to the drag force. If this force has magnitude F= 15.3v^2 find the boat speed after 12 seconds.

Homework Equations

∑F = ma
dv = adt

The Attempt at a Solution

taking the side opposite to the drag force to be positive in our force equation we get:
-15.3v² = m(-a) = m(-dv/dt).

You can multiply this equation by -1, so it is identical with 15.3v² = m(a) = m(dv/dt). That means positive acceleration and increasing speed.

so,
∫-15.3/500 dt = ∫ (1/v^2) dv with the correct limits i get V = -3.4m/s which is incorrect.

however, when i use the force equation
-15.3v² = m(a) = m(dv/dt) { no negative sign for dv/dt} i get the right answer v = 2.4 m/s .

I don't understand why this works. isn't acceleration in the same direction as the net drag force, so by how i set up my coordinate system they both should have a negative sign? Please help me understand. i looked online and in my textbook and i found nothing.

The force of drag is opposite to the velocity, it decreases the speed. Take the direction of velocity positive, then the force and acceleration are negative. .So you have to write m dv/dt= -15.3 v².

 

[sanhuy]

The force of drag is opposite to the velocity, it decreases the speed. So you have to write m dv/dt= -15.3 v2.

I understand that the velocity should decrease, but isn't acceleration in the same direction as the drag force so they should have the same signs in the force equation? This is quite non - intuitive for me.

 

[ehild]

Read the text carefully, the boat

.slows down due to the drag force. If this force has magnitude F= 15.3v^2....

So the drag force (a vector) is negative with respect to the velocity and so is the acceleration.
$\vec F_{drag} = -15.3 v^2\hat v$
when $\vec v = v \hat v$ . The equation for the acceleration is $m \frac{d\vec v}{dt} = \vec F_{drag} = -15.3 v^2 \hat v$ You can omit the vector notation writing the equation for the speed v: mdv/dt = -15.3 v².

 

[sanhuy]

Read the text carefully, the boat So the drag force (a vector) is negative with respect to the velocity and so is the acceleration.
$\vec F_{drag} = -15.3 v^2\hat v$
when $\vec v = v \hat v$ . The equation for the acceleration is $m \frac{d\vec v}{dt} = \vec F_{drag} = -15.3 v^2 \hat v$ You can omit the vector notation writing the equation for the speed v: mdv/dt = -15.3 v².

Oh I see. So, just to confirm, v hat is the unit vector that points in the direction of the velocity correct?

 

[ehild]

Oh I see. So, just to confirm, v hat is the unit vector that points in the direction of the velocity correct?

yes, $\hat v$ is the unit vector in the direction of the velocity .

 

[sanhuy]

THANK YOU! I Get it now!