course 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
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.
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?
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