Kinematics and One Dimensional Motion

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The discussion centers on whether deceleration can be assumed to be the same in different instances of one-dimensional motion. It explores the idea that the acceleration of a car during braking might be independent of its speed. A participant concludes that they have resolved their query, stating the answer is D/9. The conversation reflects a focus on the principles of kinematics and the implications of speed on deceleration. Overall, the thread emphasizes understanding the relationship between speed and braking in one-dimensional motion scenarios.
ayans2495
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Homework Statement
A car is travelling at 120 km/h, when the driver sees a herd of cows on the road ahead and slams on the brakes. The performance of the car’s brakes is such that the car comes to a stop in a distance D meters. Assuming that the acceleration of the car under braking is independent of the car’s speed, what distance would the car require to come to a stop if it were travelling at 40 km/h instead?
Relevant Equations
v=d/t, x=ut+1/2at^2
Would we assume that the deceleration of both instance are the same?
 
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ayans2495 said:
Would we assume that the deceleration of both instance are the same?
I think that's what the question is trying to say, by
Assuming that the acceleration of the car under braking is independent of the car’s speed,
 
hmmm27 said:
I think that's what the question is trying to say, by
Don't worry, I've figured it out. It's D/9. Thank you though.
 
ayans2495 said:
Don't worry, I've figured it out. It's D/9. Thank you though.
Not going to ; cheers.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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