Solve Curvilinear Motion Homework: Car Dropped from 110 ft/s in 3 secs

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A stunt car is driven off a cliff at 110 ft/s, and the problem involves calculating the gravitational acceleration normal to its path after 3 seconds. The car's horizontal distance traveled is 330 feet, while its vertical position after 3 seconds is -144.9 feet, indicating a downward motion. The velocity vector is determined to be (110, 96.6) ft/s, with the vertical component needing clarification on its direction. The discussion emphasizes the importance of understanding the normal component of acceleration, which is perpendicular to the velocity vector. The focus remains on calculating the normal acceleration using the velocity components and the principles of curvilinear motion.
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Homework Statement



A stunt car is driven off a cliff with a speed of 110 ft/s. What is the gravitational acceleration of the car normal to its path after falling for 3 seconds?

Homework Equations



The kinematic equations...?

I'm pretty sure that this should be done in Normal and tangential components, so with that said:

s = \theta r
a_t = \dot{v} = v \frac{dv}{ds} = \alpha r
v = \dot{s} = \omega r
a_n = \frac{v^2}{\rho} = \omega^2 r Where \rho is the radius of curvature.

The Attempt at a Solution



For the x-direction:
(v_0)_x = 110

t = 3

\Delta x = (v_0)_x t = (110)(3) = 330

For the y-direction:
y = y_0 + (v_0)_y t + \frac{1}{2} a t^2

Solving for distance in the y-direction:

y = \frac{1}{2}(-g)t^2 \quad (t = 3)

y = -144.9 ft


But I really have no idea if any of that is necessary, or if it is where do I go from there?
 
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Hi mathmann,

You got the position of the car, and you know that the acceleration is g, downward. You need the normal component of the acceleration. The normal of a curve is perpendicular to its tangent. And you also know that the velocity is tangent to the path at any point.

Find the velocity vector at the point (330, -144.9) first.

ehild
 
ehild said:
Hi mathmann,

You got the position of the car, and you know that the acceleration is g, downward. You need the normal component of the acceleration. The normal of a curve is perpendicular to its tangent. And you also know that the velocity is tangent to the path at any point.

Find the velocity vector at the point (330, -144.9) first.

ehild



Ok so for the x-direction since there is no acceleration then v_x = 110 and for y-direction v^2 = v_0^2 + 2(-g)(-144.9) to get v_y = 96.6, this is my guess on what to do next.

v = \sqrt{v_x^2 + v_y^2} = 146.39

But that is a scalar..? So again I'm stuck. Thank you for your first post though
 
The velocity is a vector. Write it out with its horizontal and vertical components. I see, you have set the coordinate system that the y-axis points upward.

ehild
 
ehild said:
The velocity is a vector. Write it out with its horizontal and vertical components. I see, you have set the coordinate system that the y-axis points upward.

ehild

So v = (110, 96.6) ft/s. I know this is a very basic question I just am so lost on what to do.
 
I don't know if this will be any use.. But its the picture for the question.
 

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The car moves downward. What is the sign of the y component of velocity? Could you show the velocity vector in the picture?

You need the direction perpendicular to the velocity. What do you know about the components of the vectors which are perpendicular to each other? ehild
 
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