EK 1001 Physics: Questions 57 and 65

[MCATMadness]

For 57: Particle moves at 5 m/s, reverses direction to move at 5 m/s in opposite direction. if a=constant, what is its speed at 1/2 seconds?

why cant you use -->

deltaX = VoxT +1/2AT^2
where time is 0.5s

deltaX = 5(.5) + 1/2(-10)(.5)^2
= 1.25 m

speed = distance/time
= 1.25m/0.5s

i see how the formula listed in the answer key is simpler (V = Vi + AT) but why cant you use the above formulas like you do in the earlier problems?

For 65: See page 7 for graphs

Why do graphs X and Z have constant non zero acceleration?

Acceleration = change in velocity/change in time

Wouldnt the slopes (aka velocities) change along the course of the curve --> thereby giving a non-constant non-zero acceleration? I dont see how
acceleration is constant along those curves.

This question relates to :
http://forums.studentdoctor.net/showthread.php?p=10160803#post10160803


please help!!

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[phltz]

You can only use speed = distance/time for constant velocity (which means 0 acceleration). Alternately, if you use it for an accelerating body, you get the average velocity, not the instantaneous velocity. Because they ask you for the speed at a particular moment, they want an instantaneous and not an average velocity.

I don't have the EK books, and so I dunno what to tell you about the graphs.

For 57: Particle moves at 5 m/s, reverses direction to move at 5 m/s in opposite direction. if a=constant, what is its speed at 1/2 seconds?

why cant you use -->

deltaX = VoxT +1/2AT^2
where time is 0.5s

deltaX = 5(.5) + 1/2(-10)(.5)^2
= 1.25 m

speed = distance/time
= 1.25m/0.5s

i see how the formula listed in the answer key is simpler (V = Vi + AT) but why cant you use the above formulas like you do in the earlier problems?

For 65: See page 7 for graphs

Why do graphs X and Z have constant non zero acceleration?

Acceleration = change in velocity/change in time

Wouldnt the slopes (aka velocities) change along the course of the curve --> thereby giving a non-constant non-zero acceleration? I dont see how
acceleration is constant along those curves.

This question relates to :
http://forums.studentdoctor.net/showthread.php?p=10160803#post10160803


please help!!

 

[MCATMadness]

thanks so much. that makes perfect sense now.


the graphs are just displacement vs. time graphs... with parabolic curves (one pointing up and one pointing down). basically, my question is this:

for a parabolic curve on a d vs t graph, would a be constant? or would it be changing? Page 9 of EK says the a isnt constant and therefore we cant apply LME to the curve. But then this questions says that the two parabolic graphs do have constant a... so im confused. the curves are all parabolas.

any insight?

 

[phltz]

thanks so much. that makes perfect sense now.


the graphs are just displacement vs. time graphs... with parabolic curves (one pointing up and one pointing down). basically, my question is this:

for a parabolic curve on a d vs t graph, would a be constant? or would it be changing? Page 9 of EK says the a isnt constant and therefore we cant apply LME to the curve. But then this questions says that the two parabolic graphs do have constant a... so im confused. the curves are all parabolas.

any insight?

The formula for the position of a body undergoing constant acceleration (most frequently this acceleration is due to gravity) is x = x0 + v0*t + 0.5*a*t^2. You may recognize this as an equation for a parabola. All of which is to say that, yes, a parabolic curve for displacement vs time is exactly when we have constant acceleration. So for those graphs, a is a constant.

Did you take physics with calculus? Take the second derivative of a parabola and see what you get.

 

[MCATMadness]

thanks for the clarification ... this makes sense.

what threw me off is that in EK it says that you can not apply the LME to a curved line on a d vs t graph b/c a is NOT constant.

I am guessing they meant to put that line in the v vs t graph section... a curved line on an v vs t graph would mean a is not constant and therefore, the LME eqns do not apply.

does this sound right? thnx for your help!

 

[Rabolisk]

A curved line on a d vs t graph only shows that v is not constant. Acceleration can be constant if the graph is a parabola, but not all curved lines are necessarily parabolas. A curved line on a v vs t graph would mean that a is not constant.

 

[MCATMadness]

got it thanks!