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

In summary, the conversation is discussing a physics problem involving a stunt car driven off a cliff with a speed of 110 ft/s. The question is asking for the gravitational acceleration of the car normal to its path after falling for 3 seconds. The conversation also mentions the kinematic equations and the need to find the normal and tangential components of acceleration. The solution involves finding the position and velocity vectors of the car, with the y-component of velocity being negative since the car is moving downward. The final step is to determine the perpendicular direction to the velocity vector, using the fact that perpendicular vectors have components that are negative reciprocals of each other.
  • #1
mathmannn
15
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:

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

The Attempt at a Solution



For the x-direction:
[itex] (v_0)_x = 110 [/itex]

[itex] t = 3 [/itex]

[itex] \Delta x = (v_0)_x t = (110)(3) = 330 [/itex]

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

Solving for distance in the y-direction:

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

[itex] y = -144.9 ft [/itex]


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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  • #2
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
 
  • #3
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 [itex] v_x = 110 [/itex] and for y-direction [itex] v^2 = v_0^2 + 2(-g)(-144.9) [/itex] to get [itex] v_y = 96.6 [/itex], this is my guess on what to do next.

[itex] v = \sqrt{v_x^2 + v_y^2} = 146.39 [/itex]

But that is a scalar..? So again I'm stuck. Thank you for your first post though
 
  • #4
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
 
  • #5
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 [itex] v = (110, 96.6) [/itex] ft/s. I know this is a very basic question I just am so lost on what to do.
 
  • #6
I don't know if this will be any use.. But its the picture for the question.
 

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  • #7
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:

1. How do you calculate the velocity of the car after 3 seconds?

To calculate the velocity of the car after 3 seconds, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity (110 ft/s in this case), a is the acceleration (due to gravity, which is 32 ft/s^2), and t is the time (3 seconds). Plugging in the values, we get v = 110 ft/s + (32 ft/s^2)(3 s) = 206 ft/s.

2. What is the distance traveled by the car in 3 seconds?

The distance traveled by the car in 3 seconds can be found by using the formula s = ut + 0.5at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get s = (110 ft/s)(3 s) + 0.5(32 ft/s^2)(3 s)^2 = 330 ft + 144 ft = 474 ft.

3. How long does it take for the car to hit the ground?

We can use the formula t = (v-u)/a to calculate the time it takes for the car to hit the ground, where t is the time, v is the final velocity (which will be 0 when the car hits the ground), u is the initial velocity, and a is the acceleration. Plugging in the values, we get t = (0 ft/s - 110 ft/s)/-32 ft/s^2 = 3.44 seconds. So, it takes approximately 3.44 seconds for the car to hit the ground.

4. What is the acceleration of the car?

The acceleration of the car is the rate at which its velocity changes. In this case, the car is accelerating due to gravity at a rate of 32 ft/s^2. This means that every second, the car's velocity will increase by 32 ft/s. So, the acceleration of the car is 32 ft/s^2.

5. Can this same process be used to calculate the motion of objects in other scenarios?

Yes, this same process can be used to calculate the motion of objects in other scenarios. The equations used in this example are part of the equations of motion, which are fundamental equations in physics that describe the motion of objects in different scenarios. These equations can be used to analyze the motion of objects in various situations, such as free fall, projectile motion, and circular motion.

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