Uniform Circular Motion and Centripetal acceleration

[suyashr99]

Homework Statement

For a scene in a movie, a stunt driver drives a 1.20✕ 103 kg pickup truck with a length of 4.65 m around a circular curve with a radius of curvature of 0.333 km (Fig. 7.29). The truck is to curve off the road, jump across a gully 10.0 m wide, and land on the other side 2.96 m below the initial side. What is the minimum centripetal acceleration the truck must have in going around the circular curve to clear the gully and land on the other side?

Homework Equations

a(c) = v^2/ r
Fc = mv^2/ r
(I don't really know what else to apply here)
I guess we could use energy or kinematics for the ravine part.
KE= 1/2mv^2
PE= mgh

The Attempt at a Solution

0.333 km= 333 m
Fc = mv^2/ r
This is where i got lost and confused.
Please guide me.

 

[suyashr99]

Attempt 1: Used the maximum height that it can travel to find the time it would take to cross the gulch (t = 0.77 secs). Divided that by distance to get a velocity (12.9). Plugged that into the formula for centripetal acceleration and got .49 which is wrong.

 

[Mister T]

I think you have to take into account the length of the truck, otherwise you are calculating the speed it would take for the truck's front bumper to clear the gully, leaving the rest of the truck to fall in!

 

[suyashr99]

I think you have to take into account the length of the truck, otherwise you are calculating the speed it would take for the truck's front bumper to clear the gully, leaving the rest of the truck to fall in!

Okay so i woulld just add right? thanks ill try that.

 

[suyashr99]

I think you have to take into account the length of the truck, otherwise you are calculating the speed it would take for the truck's front bumper to clear the gully, leaving the rest of the truck to fall in!

Okay thank you that was correct

 

[SWJ]

Hi, i am really interested in this question and would like to attempt this question on my own. So is it possible if you can post the answer to this question? Thank you

 

[haruspex]

Hi, i am really interested in this question and would like to attempt this question on my own. So is it possible if you can post the answer to this question? Thank you

No, you should post your own attempt first.

 

[SWJ]

No, i meant the final answer to this question, not the solution so that i know if what i get in the end is correct.

 

[suyashr99]

The answer is 1.08 m/s^2