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Question that seems simple but I cant fully get it

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A rock is thrown straight up into the air at a speed of 25m/s. After a period of 7s

a) Find the change in displacement of the rock. (65m [down])
b) Find the total distance the rock has travelled (130m)
c) How long does it take to get to the top? (2.6s)
I was able to figure out part c (by using -9.8 and the acceleration forumla), comes out to 2.55s, rounded to 2.6s. No problems there.

For part a, I used Velocity average = delta d / delta t
(25+0)/2 = delta d / (2.6*2) I did get the right answer, but I dont fully understand the conceptual part if it, I keep thinking it should be 0, since it goes straight up and straight back down to starting point? Also why do you need a direction for the answer?

As for b), I never got to it, but I suppose for some reason, you need to double the answer for a to get it...

Lastly, I think the 7.0s is just there to throw people off?
 
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For Part a)
You aren't looking for a velocity, you are looking for a displacement. Theres a different equation you are looking for. Also to clarify the question, it is asking for how far the rock is from the starting point 7 seconds after its thrown.

b) This one is similar to a, except its asking how much distance it has covered total, not just how far it is from where it started ( at t = 7 sec).

I don't understand how it could get to teh top at t=2.6s if its asking for information at t=7 sec
 
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Displacement is a vector, so it needs a direction.
As for the distance, well, if the rock travels 65 m upwards and then comes all the way back down, of course it will have travelled a total distance of 130 m.
Well no, the question is asking you to find those quantities in those 7 seconds..
 
after the rock is thrown the only force acting on it is gravity, and so the acceleration of the rock is constant.
you should have some nice equations for constant acceleration that can thus be applied.

for part a) you can use [itex]s - s_{0} = v_{0}t + \frac{1}{2}at^2[/itex], but be careful that your initial velocity ([itex]v_{0}[/itex]) and your acceleration are in opposite directions (i.e have opposite signs). you'll have to decide which way is positive (usually it's up, but putting it as down can sometimes be helpful). the sign of your answer should then tell you which direction it is from the start. you need to interpret this sign as a direction and quote it in your answer. displacement is a vector quantity and therefoe has a magnitude and direction

for part b), the displacement at the top of the rock's trajectory can be found. you know the "final" velocity (for this part at least), which is momentarily zero. You want displacement, so [itex] v^2 = v_{0}^2 + 2a(s - s_{0})[/itex] should help you. It may be helpful to imagine that the rock is thrown from the top of a cliff, and goes over the edge on it's way down, and then calculate the total distance. drawing a picture of these types of problems is often a neccessity!

for part c) if you were referring to [itex]v = v_{0} + at[/itex] (which i think you were) then you were spot on :smile:
 
Last edited:

HallsofIvy

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I have a question about part (a): Is the rock thrown upward with your hand out over the edge of a cliff??

If you were standing on a flat surface then, as said before, the rock goes up for 2.6 s, then comes back down for 2.6 seconds and hits the ground! After it lies there for another 7- 4.2= 2.8 seconds, it's still on the ground so the total displacement is 0.

If, however, you throw the rock upward on the edge of a cliff, the rock goes up for 2.6 seconds then back downward for 2.6 seconds (at that point its height is 0 and its speed is 25 m/s downward). For the final 4.2 seconds, it is accelerating downward so the distance it goes in that time is d= -4.9(4.2)2- 25(4.2)= -190 m (I'v rounded of to two significant figures as given in the problem). Since the problem asks for "displacement", in this case you could say either "190 m downward" or "-190 m".
 

VietDao29

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HallsofIvy said:
...If you were standing on a flat surface then, as said before, the rock goes up for 2.6 s, then comes back down for 2.6 seconds and hits the ground! After it lies there for another 7- 4.2= 2.8 seconds, it's still on the ground so the total displacement is 0.

If, however, you throw the rock upward on the edge of a cliff, the rock goes up for 2.6 seconds then back downward for 2.6 seconds (at that point its height is 0 and its speed is 25 m/s downward). For the final 4.2 seconds, it is accelerating downward so the distance it goes in that time is d= -4.9(4.2)2- 25(4.2)= -190 m (I'v rounded of to two significant figures as given in the problem). Since the problem asks for "displacement", in this case you could say either "190 m downward" or "-190 m".
There is something wrong... 2.6 x 2 = 4.2?? :confused:
VietDao,
 
that is not the only mistake.

also, i only brought in the cliff analogy to illustrate an example of how negative displacement could occur in this problem. as an answer of "65m [down]" was given, it is clearr that the rock does not fall to the ground.

it is okay too say -65m as long as you have stated somewhere (like clearly on a diagram) that up is positive. that's what i was taught anyway.
 

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