Calculating Acceleration of a Toy Plane Flying in a Circle

AI Thread Summary
The discussion centers on calculating the acceleration of a toy plane flying in a circle with a radius of 2 meters, completing three laps in three seconds. The initial calculations for distance and tangential velocity are confirmed, leading to an acceleration of approximately 78.96 m/s² using the formula a = v²/r. However, it is noted that the problem involves constant tangential acceleration, which requires a different approach. The conversation emphasizes the need to consider both tangential and normal acceleration components to determine the overall acceleration vector. Ultimately, the focus remains on accurately computing the magnitude of the acceleration.
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Question: Starting from rest, the toy plane flies around a circle of radius 2m, three times in 3 seconds. There is constant tangential acceleration, fins the magnitude of the acceleration at the end of 0.5s.

My solution (most likely wrong):

Circumference= 2pir= 4pi= 12.56631

Distance traveled in 3 rounds (or in 3 seconds )

= 3* circumference= 37.699m.

At 0.5 seconds (1/6th of the total time), distance travelled

= 37.699/6= 6.2831m.

Velocity at 0.5s

= displacement/time = 6.2831m/0.5s= 12.56631 m/s.

( I knew I didn’t have to do all those steps but I did to help myself)

Acceleration at 0.5s

a= v^2/r = (12.56631)^2/2 = 78.96 m/s^2.

Am I okay? Or did I make a mistake, if so, where did I go wrong?

Thanks in advance for the help!
 
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It seems okay to me. You were a little bit roundabout to get there though.

If it takes 3 secs to travel around 3 times then that's once per second and the circumference is 12.56631 m hence 12.56631 m/s is the speed (tangential velocity) and from there you can compute the acceleration as a = v^2/r = (12.56631)^2/2 = 78.956 m/s^2

so we agree.
 
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jedishrfu said:
It seems okay to me. You were a little bit roundabout to get there though.

If it takes 3 secs to travel around 3 times then that's once per second and the circumference is 12.56631 m hence 12.56631 m/s is the speed and from there you can compute the acceleration as a = (12.56631)^2/2 = 78.956 m/s^2

Oh phew, thanks!
 
Oops I didnt see the constant tangential acceleration. What we solved was for constant tangential velocity.

I think that means it took 3 seconds to do three orbits so if you were to stretch the circle out to a straight line then and use the s = 1/2 a t^2 and find the a
 
jedishrfu said:
Oops I didnt see the constant tangential acceleration. What we solved was for constant tangential velocity.

I think that means it took 3 seconds to do three orbits so if you were to stretch the circle out to a straight line then and use the s = 1/2 a t^2 and find the a

So my s would just be the distance traveled in 3 orbits, and this also means that the magnitude of the plane’a velocity changes throughout the course of the 3 orbits?
 
The next question would be at 0.5 sec there is a tangential and a normal acceleration component to construct the acceleration vector. You’ll need the velocity at 0.5 sec to compute the normal acceleration via v^2 / r

So you could use Pythagorean theorem to get the acceleration magnitude.

How far do you need to go? To get the actual vector?

Your problem says you need only compute the acceleration magnitude.
 
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