Question

Assume that the car at point A and the one at point E are traveling along circular paths that have the same radius. If the car at point A now moves twice as fast as the car at point E, how is the magnitude of its acceleration related to that of car E?

Answer 1

Answer:

The magnitude of the acceleration of the car at point A is four times that of the car at point E

Explanation:

Car at point A and at point E are traveling along the same radius

So, they both have a common radius r

Now, car At point A now move with twice the velocity of car at point E

Let the velocity of car at point E be Ve=V

Then, the velocity of the car at point A is twice that at point E,

Therefore Va =2V

Now we want to compare their acceleration

The centripetal acceleration of a circular motion is given as

a=v²/r

Then, for car at point E, since Ve=V, the acceleration will be,

ae=v²/r

ae=V²/r

Then, for car at point A, since Va=2V, the acceleration will be

aa=v²/r

aa=(2V)²/r

aa=4V²/r

aa=4•V²/r. Since ae=V²/r

aa=4•ae

Note: ae is acceleration at point E and aa is the acceleration at point A

It is notice that the acceleration of the car at point A is four time that of point E.

Answer 2

Explanation:

Below is an attachment containing the solution.