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

  • Thread starter Thread starter mathmannn
  • Start date Start date
  • Tags Tags
    Curvilinear Motion
Click For Summary
SUMMARY

The discussion focuses on calculating the normal component of gravitational acceleration for a stunt car dropped from a height with an initial speed of 110 ft/s after 3 seconds. The participant correctly identifies the need to separate the motion into normal and tangential components, utilizing kinematic equations. The final velocity vector is determined to be (110, 96.6) ft/s, leading to the calculation of the normal acceleration component, which is essential for understanding curvilinear motion.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Familiarity with vector components and their representations
  • Knowledge of gravitational acceleration (g = 32.2 ft/s²)
  • Concept of normal and tangential acceleration in curvilinear motion
NEXT STEPS
  • Study the derivation of normal and tangential acceleration components in curvilinear motion
  • Learn how to apply kinematic equations to solve for velocity and acceleration in two dimensions
  • Explore the concept of radius of curvature and its impact on motion
  • Investigate graphical representation of velocity vectors in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and curvilinear motion, as well as educators seeking to clarify concepts related to acceleration and velocity vectors.

mathmannn
Messages
15
Reaction score
0

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?
 
Physics news on Phys.org
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.
 

Attachments

  • Screen Shot 2012-01-28 at 12.39.04 AM.png
    Screen Shot 2012-01-28 at 12.39.04 AM.png
    2 KB · Views: 493
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
 
Last edited:

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
6
Views
2K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
13
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
Replies
7
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
6
Views
1K