Question

How would we find the velocity of the particle as it moves through the point (9/3,3)?

Answer 1

The velocity at the given point is  (16 units/second)*( 1/√5,   1/2√5)

Where (1/√5,   1/2√5) defines the direction.

How to find the particle's velocity?

We know that the particle moves along the top branch of:

y^2 = 3x

Then we can rewrite that equation as:

y = √(3x)

Because the particle moves along the top branch, we only look at the positive square root.

Now, the point (9/3, 3) clearly belongs to the curve, we can see that by evaluating in: x = 9/3.

y =  √(3*(9/3)) =  √9 = 3

Now we know that the speed of the particle is 8 units per second, and we want to find the velocity. Remember that the velocity also has a direction, so the direction needs to be along the curve.

To find the direction, we can find the slope of the tangent line at that point by derivating and evaluating in x = 9/3.

The first derivation is:

y' = (3/2)*(3x)^{-1/2}

Evaluating in x = 9/3:

y' = (3/2)*(3*(9/3))^{-1/2} = (3/2)*(3)^-1 = (1/2)

So the slope of the tangent line at that point is 1/2. And the velocity's direction is given by that line.

Then the direction, at that point, is something of the form: (x, 1/2*x)

Taking the variable as an unit x = 1 (we want to find a unit vector that defines the direction) we get: (1, 1/2)

Now we need to normalize that.

The magnitude of that vector is:

M = √(1^2 + (1/2)^2) = √(1 + 1/4) = √(5/4) = √5/2

Then the unit vector is equal to the vector we found above divided by √5/2, this is:

(2/√5)*(1, 1/2)

That defines the direction of our velocity, and we know that the speed (the magnitude of the velocity) is 8 units per second.

Then the velocity is:

V = (8 units/second)*(2/√5)*(1, 1/2)

   = (16 units/second)*( 1/√5,   1/2√5)

If you want to learn more about velocity and speed:

brainly.com/question/4931057

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