#26 from 1001 examkrackers physics

[realestate]

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This is from examkrackers 1001 physics. really rusty with physics so probably a newb question.

"A particle moves along path C at a constant speed of 1m/s. What is the average acceleration of the particle as it moves from position 1 to position 2?"

A. 0
B. 0.2/pi
C. 0.4/pi
D. 1

The book says C is correct. Acceleration is change in velocity divided by time. Initial velocity is 1m/s up; final velocity is 1m/s down. The change in velocity is therefore 2m/s. The time is found from speed equals distance divided by time. Distance is 2pir/2. Thus a= 2/[(2pi5/2)/1]. 2/(5pi) = 0.4/pi

how do you get the average acceleration though? thanks

 

[PiBond]

This is from examkrackers 1001 physics. really rusty with physics so probably a newb question.

"A particle moves along path C at a constant speed of 1m/s. What is the average acceleration of the particle as it moves from position 1 to position 2?"

A. 0
B. 0.2/pi
C. 0.4/pi
D. 1

The book says C is correct. Acceleration is change in velocity divided by time. Initial velocity is 1m/s up; final velocity is 1m/s down. The change in velocity is therefore 2m/s. The time is found from speed equals distance divided by time. Distance is 2pir/2. Thus a= 2/[(2pi5/2)/1]. 2/(5pi) = 0.4/pi

how do you get the average acceleration though? thanks

average acceleration is just the change in velocity over the elapsed time. what makes this question tricky is that time is not given so we must solve for it using v=d/t. The distance is just half of a circle (2piR/2) and we're given velocity.

 

[JmanDO]

And don't forget, velocity is a vector, which means direction counts (as well as magnitude)! If the numerical value stays the same but the direction changes, the velocity is also changing!

I found lots of people tried to pull that trick on me, at least in the beginning.

 

[mzblue]

Same diagram as above. The question states:
A particle moving from position 1 to position 2 moves along path C. it travels at a constant speed of 5pi m/s. At exactly half way the trip its average vertical velocity is

a. 0/m/s
b. 0.5m/s
c. 1m/s
d. 10m/s

here's what i thought, since it's a circle, i thought at halfway through the trip, we'll be at max height so vertical velocity will be zero/ but then a is wrong.
The explanation points out that half way though the trip, the vertical displacement is 5. Agreed, it'll be the same as the radius. Then it states the time for half the trip is given by speed = distance/time where distance is 2pir/4.

what i don't get is, why isn't the vertical velocity at the top zero? is it because it states it's half a circle? Does it have to say "parabola" before we assume velocity at halfway is zero? I'm a little confused?

 

[ksmi117]

Same diagram as above. The question states:
A particle moving from position 1 to position 2 moves along path C. it travels at a constant speed of 5pi m/s. At exactly half way the trip its average vertical velocity is

a. 0/m/s
b. 0.5m/s
c. 1m/s
d. 10m/s

here's what i thought, since it's a circle, i thought at halfway through the trip, we'll be at max height so vertical velocity will be zero/ but then a is wrong.
The explanation points out that half way though the trip, the vertical displacement is 5. Agreed, it'll be the same as the radius. Then it states the time for half the trip is given by speed = distance/time where distance is 2pir/4.

what i don't get is, why isn't the vertical velocity at the top zero? is it because it states it's half a circle? Does it have to say "parabola" before we assume velocity at halfway is zero? I'm a little confused?

You are right that the instantaneous vertical velocity at the top is zero, but that's not what the question asked for. It asked for the average vertical velocity. This will be equal to the (total vertical displacement)/(total time).

So find out how much time using the same method below. We are travelling around 1/4 of the circle, so d=2pi*r/4= pi*r/2 = 5pi/2. So it takes .5 sec to travel this far (using the given velocity in problem).

Thus, my average vertical velocity = 5/.5 = 10 m/s.

 

[crafty27]

I am having difficulty with finding the change in velocity. The explanation given does not make sense to me. How do you get a change in velocity of 2m/s?

 

[mcloaf]

I am having difficulty with finding the change in velocity. The explanation given does not make sense to me. How do you get a change in velocity of 2m/s?

The initial velocity at the moment the particle leaves the first position is 1m/s straight up, which we are counting as + 1 m/s. The final velocity as it reaches the second position is 1m/s straight down. Since we assigned the upwards direction positive values, the downward direction must have negative values, which means our final velocity is -1 m/s. The change in velocity is the difference between the two: +1m/s - -1m/s = 1m/s + 1m/s = 2m/s.

 

[straight2thapros]

I don't get the answer for #26!! Can someone explain why we couldn't just use v^2/r?? Is average acceleration different than centripetal acceleration?

Thanks!

 

[DrknoSDN]

I don't get the answer for #26!! Can someone explain why we couldn't just use v^2/r?? Is average acceleration different than centripetal acceleration?

Thanks!

This question is asking about the vector average sum, so here there is a difference.

If you looked at an object in orbit there is a constant centripetal acceleration at any point you measure. That is different than taking the vector sum of all the instantaneous accelerations and averaging them. If an object traveled in a circle and you measured the acceleration vectors at 0, 90, 180, and 270 degrees, then took the average sum it would be zero. If you measured it an infinite number of times it would still average out to zero for a circle. That is different than taking all the positive acceleration scalar values and averaging it out to get a positive number.

For this question you have a semi circle so the horizontal components of the vectors perfectly cancel each other out. The average acceleration is then just the sum of the vertical components of all the vectors. They solved by using total change in vertical velocity over time.