course Phy 201
Below are my corrections for random assignment 3
random_assgn_3course Phy 201
What are the acceleration and final velocity of an object whose average velocity, accelerating uniformly from rest for 7.9 seconds, is 59 cm/s? How far does the object travel during the 7.9 seconds?
You should start by clearly identifying the given information.
7.9 seconds is the time interval `dt
59 cm/s is the average velocity vAve
The object starts from rest so its initial velocity is v0 = 0.
Average acceleration = change of velocity/ time elapsed so acceleration = 59 cm/7.9sec = 7.468 cm/sec
&&&& change in velocity is 59 cm/s - 0 / time elapsed = 7.9 sec = 59cm/s/ 7.9sec = 7.468 cm/s^s&&&&
The units are good on your revision.
However 59 cm/s is not the final velocity, it is the average velocity. Accel is uniform and init vel is zero so final vel = 2 * ave vel = 118 cm/s.
This will double your result for the average acceleration.
Right definition, right type of calculation, but be sure you get the units right. There is no such thing in this problem as 59 cm; there is a quantity which is 59 cm/s, and if you use this quantity your calculation would be
59 cm/s / 7.9sec = 7.468 cm/sec^s.
However 59 cm/s is the average velocity, not the change in velocity, so this still wouldn't quite fit the definition.
If initial velocity is 0 and average veloctiy is 59 cm/s, what must be the final velocity? That is, what velocity would you average with 0 cm/s to get 59 cm/s?
And the final velocity would be 59 + 7.468 = 66.468 cm/sec.
&&&& final velocity = initial velocity + acceleration(time) so vf = 0 + 7.468 cm/s^s(7.9 sec) = 59cm/s&&&&
That would be right if the accel. was 7.468 cm/s^2; however this is circular since the 7.468 cm/s^2 was based on the premise that the final velocity is 59 cm/s.
59 and 7.468 are numbers corresponding to different quantities, which cannot be added. With units your calculation would be
59 cm/s + 7.468 cm/s^2.
The two quantities have different units and hence cannot be added as 'like terms'.
59 cm/s is
And the distance traveled = time * average speed so distance traveled = 7.9 * 7.468 = 58.997 cm.
Right definition, but you're missing units on the left-hand side. The units of 7.468 are cm/s^2, and that quantity is not an average velocity.
What is the average velocity, what is the time interval, and what therefore is the displacement?
&&&& distance traveled = time * average speed = 7.9sec * 59 cm/s = 466.1cm&&&&
this is correct
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end). &#
You missed the point about final velocity and average velocity.
No need to submit a revision unless you're unsure of something, but see my notes and be sure to let me know if you have questions.