# Pytels Dynamics 12.10: parabolic path, velocity, acceleration

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1. Jun 12, 2017

### Alexanddros81

1. The problem statement, all variables and given/known data
An automobile goes down a hill that has the parabolic cross section shown. (see image attached)
Assuming that the horizontal component of the velocity vector has a constant
magnitude v0, determine (a) the expression for the speed of the automobile in
terms of x; and (b) the magnitude and direction of the acceleration.

2. Relevant equations
y=h(1-x2/b2)

3. The attempt at a solution
I have attached an image

It gives a solution in (a) v0√1+(2hx/b2)2
and (b) 2hv02/b2

How do I proceed in (a) to come to the given solution?

#### Attached Files:

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• ###### Pytel_Dynamics_12_10b.jpg
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2. Jun 12, 2017

### kuruman

Have you considered mechanical energy conservation?

3. Jun 12, 2017

### Alexanddros81

Conservation of mechanical energy is introduced later in pages 148-149.
This problem is at the first pages(pg. 23) of Dynamics 2nd edition under the Rectangular Coordinates section.

4. Jun 12, 2017

### kuruman

Actually, I was off the mark initially. This is not an energy conservation problem, but a simple application of calculus. You are looking for $v = \sqrt{v_x^2+v_y^2}$. You know that $v_x = v_0$, that $y = h(1-x^2/b^2)$ and that $v_y = dy/dt$. So ...

5. Jun 12, 2017

### Alexanddros81

so for vy = dy/dt i get vy = - (h/b2)2x

and v = sqrt(v02 + (- (h/b2)2x)2)

I don't know how to proceed then

6. Jun 12, 2017

### kuruman

You did not apply the chain rule of differentiation carefully enough. Try again. Remember that x depends on time.

7. Jun 16, 2017

### Alexanddros81

Ok. What I get is different from the given soution at the back of the book.
It is given $v = v_0 \sqrt {1 + \left( \frac {2hx} {b^2}\right) ^2}$
what I get is $v = v_0 \sqrt {1 + \left( \frac {-2hx} {b^2}\right) ^2}$

#### Attached Files:

• ###### Pytel_Dynamics014.jpg
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8. Jun 16, 2017

### kuruman

$\left( -\frac{2hx}{b^2} \right)^2=\left( (-1)\frac{2hx}{b^2} \right)^2=(-1)^2\left( \frac{2hx}{b^2} \right)^2=\left( \frac{2hx}{b^2} \right)^2$

9. Jun 16, 2017

### Alexanddros81

Any hint in finding the magnitude of acceleration?
It should be $\frac {2hv_0^2} {b^2}$

I have done the $a_y=\frac {d} {dt}\left( v_y\right) = \frac {d} {dt} \left [v_0\left(\frac {-2hx} {b^2}\right)\right]$. How do i proceed from here?

and $a_x$ should be zero since $v_x$ is constant

10. Jun 16, 2017

### ehild

Apply Chain Rule, (see @kuruman's Post#6)

11. Jun 16, 2017

### Alexanddros81

It should be as shown at the attached file

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