A train slows down at a constant rate as it rounds a sharp circular horizontal turn. Its initial speed is not known. It takes 18.1 s to slow down from 80 km/h to 33 km/h. The radius of the curve is 163 m. As the train goes around the turn, what is the magnitude of the tangential component of the acceleration? Answer in units of m/s2

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

A train slows down at a constant rate as it rounds a sharp circular horizontal turn. Its initial speed is not known. It takes 18.1 s to slow down from 80 km/h to 33 km/h. The radius of the curve is 163 m. As the train goes around the turn, what is the magnitude of the tangential component of the acceleration? Answer in units of m/s2

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

So its represented by a change in distance over time.... therefore 80-(33km/hr)/18.1 s thn convert because its asking for m/s^2
(13.06m/s)/18.1s = .721 ms^-2
Then As the train goes around the turn, what is the sign of the tangential component of the acceleration? Take the direction in which the train is moving to be the positive direction. I just use the sign of the previous answer which is positive? right?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

1) this is not a question of distance 2) you can't just divide by time again to get m/s^2, you need to follow the physics all acceleration is in m/s^2 in SI units the formula you need here seems to be\[a_{tan}=\frac{\Delta v}{\Delta t}\]
...not sure why the radius of the curve is given
note: you will need to convert km/hr into m/s
isnt the velocity 80km/hr - 33km/hr ????
47 km/hr converts to 13.06 m/s
that is delta v, but you said that was distance in your answer and yes to your conversion, so I guess you had the right answer, you just phrased it strangely for me looks good now though :D
I just spoke incorrectly... thankyou
what about the second part of the question
is it positive?
It will be negative because it slows the train.

Not the answer you are looking for?

Search for more explanations.

Ask your own question