Tbr physics error example?

[crazy person]

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For chapter 1, on page 20, when they said they arrived at the equation y= 1/2at by using equation v= vo+at, isn't this wrong? shouldn't it be from equation y= vot + 1/2 at^2?

on page 23, they also said, using equation 1.18 which is y=1/2at^2 ? when it should be v=vo +at, right? and the answer should be -4s not 4s, right?

I think there is an error they made or am I wrong?

 

[AFLATPEG]

For chapter 1, on page 20, when they said they arrived at the equation y= 1/2at by using equation v= vo+at, isn't this wrong? shouldn't it be from equation y= vot + 1/2 at^2?

on page 23, they also said, using equation 1.18 which is y=1/2at^2 ? when it should be v=vo +at, right? and the answer should be -4s not 4s, right?

I think there is an error they made or am I wrong?

yea it's just a typo on TBR's part.

 

[BerkReviewTeach]

For chapter 1, on page 20, when they said they arrived at the equation y= 1/2at by using equation v= vo+at, isn't this wrong? shouldn't it be from equation y= vot + 1/2 at^2?

You are absolutely 100% correct that it's a typo. It should have said, "We arrived at this expression by using equation (1.14) and setting y_o = 0, v_o = 0, and t_o = 0" as opposed to saying "We arrived at this expression by using equation (1.15) and setting y_o = 0, v_o = 0, and t_o = 0"

on page 23, they also said, using equation 1.18 which is y=1/2at^2 ? when it should be v=vo +at, right? and the answer should be -4s not 4s, right?

I think there is an error they made or am I wrong?

Think about this conceptually. If a ball is tossed upwards it has +v_y and is slowing down as it climbs. The point at which it reaches the top of its flight will be the point where it has v_y = 0. Does this occur before or after you first throw the ball? The answer must be after, so it is occurring at a later time, making the time +4 s. To be -4 s would require going back in time.

In terms of math, given that v_y = 0 at the top of the flight, we get:

0 = v_oy + gt​

so:

v_oy = -gt​

given that g = -9.8 m/s² and v_oy = +40 m/s, we get:

40 = -(-9.8)t

40 = 9.8t​

so:

t = 40/9.8 = 4 + a little​

 

[mehc012]

Minor quibble, since we're going for clarity on small points here: I want to point out that 'a' = -g, because g = 9.8m/s² and is actually just the MAGNITUDE of acceleration due to gravity. Some eq'ns will expect you to use 'g' by itself, which is the only reason it matters at all.

For the above,
v_0y = -at_____a = -g
v_0y = -(-g)t
v_0y = gt_____g = 9.8 m/s² (no negative sign)

 

[BerkReviewTeach]

mehc012, I'm not sure what the minor quibble here is. The sign convention applied to this question is that upward is (+) and downward is (-), so the initial velocity (v_o) is + 40 m/s and the acceleration (a = g) is - 9.8 m/s². Are you emphasizing that the sign should be eliminated before plugging 9.8 in rather than after?

What I typed and you typed look to be essentially the same thing to me, except for my emphasis on following a reference frame for (+) and (-) directions from the start of the calculation.

v_oy = -at
v_oy = -gt
+40 = -(-9.8)t
+40 = 9.8t
⁴⁰/_9.8 = t​

versus

v_oy = -at
v_oy = -(-g)t
+40 = gt
+40 = 9.8t
⁴⁰/_9.8 = t​

From a purely MCAT-driven perspective, the sign for time in this case should be based on the conceptual notion that the top of the flight happens after it is launched, so it must be a + value for time. Haggling over the signs being plugged into the calculation doesn't help in terms of how to solve MCAT questions like this, and at least in my opinion creates an uneccesary extra layer of concern that actually takes away from what someone should be doing to solve this question efficiently.

 

[mehc012]

I'm sorry, it was super, SUPER minor, to the point of being irrelevant, and I shouldn't have posted anything.
I just was taught that g was the magnitude of acceleration due to gravity, and is always positive. I am also keeping the sign convention, I'm just specifying that a = -g in order to show it. I find it easier to keep track of this way, because there are some situations in which positive g should be used, and by keeping the sign of 'g' consistent, I avoid getting confused with signs in those situations.

 

[crazy person]

you are absolutely 100% correct that it's a typo. It should have said, "we arrived at this expression by using equation (1.14) and setting y_o = 0, v_o = 0, and t_o = 0" as opposed to saying "we arrived at this expression by using equation (1.15) and setting y_o = 0, v_o = 0, and t_o = 0"



think about this conceptually. If a ball is tossed upwards it has +v_y and is slowing down as it climbs. The point at which it reaches the top of its flight will be the point where it has v_y = 0. Does this occur before or after you first throw the ball? The answer must be after, so it is occurring at a later time, making the time +4 s. To be -4 s would require going back in time.

In terms of math, given that v_y = 0 at the top of the flight, we get:

0 = v_oy + gt​

so:

v_oy = -gt​

given that g = -9.8 m/s² and v_oy = +40 m/s, we get:

40 = -(-9.8)t

40 = 9.8t​

so:

t = 40/9.8 = 4 + a little​

thank u ! 🙂

 

[Landon911]

I remember lots of typos in my tbr books. My personal favorite was the claim oxygen in water was REDUCED to oxygen gas. I will never forget that one.