How to find the final velocity o

AI Thread Summary
To find the final velocity of a bullet hitting a bulletproof vest, the equation Vf^(2) = Vi^(2) + 2ad is suggested, where Vi is the initial velocity of 80 m/s, a is the deceleration of 9000 m/s², and d is the thickness of the vest at 0.26 m. The discussion highlights confusion over whether the deceleration is constant over the entire distance of 6.26 m. Participants emphasize the importance of showing calculations to identify errors in the approach. Clarification on the constant nature of acceleration is also sought, indicating a need for more detailed problem-solving steps.
Jessica Lewis
Messages
3
Reaction score
0
The question is:
a bullet that is shot from a target is traveling at an initial speed of 80 m/s and it is 6 m away from its target. The target is wearing a bullet proof vest that is 0.26 m thick. As the bullet makes contact with the vest is slows down at a rate of 9000 m/s2. What is the final velocity of the equation?I would assume that the equation used would be Vf^(2) = Vi^(2) + 2ad, however each time I attempt this question I get an answer that makes absolutely no sense.
So, please help me,
Thank you! :)
 
Physics news on Phys.org
EDIT: I realized I did the problem and, according to the guidelines, that isn't allowed! So, instead: for the acceleration, is 9000 m/s2 constant for the whole 6.26 m?
 
Last edited:
Jessica Lewis said:
each time I attempt this question I get an answer that makes absolutely no sense.
So, please help me,
Thank you! :)
how do you expect anyone to tell you where you are going wrong if you do not post your working?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top