Final velocity given acceleration and initial velocity

  • Thread starter Ohoneo
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  • #1
Ohoneo
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Homework Statement


A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.1 s later it is rising at a speed of 19 m/s. Assuming air resistance has no effect on the rock, calculate its speed at (a) at launch and (b) 5.2 s after the launch.


Homework Equations


v = v0 + at (I'm assuming is the only relevant one, although I'll post two others in case they are needed)
x = x0 + v0t + a/2 t^2
v^2 = v0*2 + 2a(x-x0)


The Attempt at a Solution



I solved part A by plugging into the equation v = v0 + at
19 m/s = v0 + (-9.8 m/s^2)(2.1 s)
And I found that the initial velocity equals 40 m/s.

So, to solve part b, I should just have to plug the initial in and find the final. I tried that:
v = 40 m/s + (-9.8 m/s^2)(5.2 s)
v = -11 m/s

However, when I entered that solution in for the homework, I was told it was wrong. I'm not really sure how to go about doing the problem if that's incorrect. I thought maybe I could have an error in rounding with significant figures.

Thanks for the help!
 

Answers and Replies

  • #2
gneill
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It could be that since part b asked for the speed, not the velocity, that the sign of the number reported should have been positive.
 
  • #3
p21bass
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The problem asked for the speed of the rock - you gave its velocity. Remember - speed is a scalar (independent of direction), and velocity is a vector (depends upon direction).


edit: gneill and I were posting at the same time, apparently.
 
  • #4
Ohoneo
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The problem asked for the speed of the rock - you gave its velocity. Remember - speed is a scalar (independent of direction), and velocity is a vector (depends upon direction).


edit: gneill and I were posting at the same time, apparently.

I tried that as the answer (11 m/s instead of -11 m/s) and it was still incorrect.
 
  • #5
p21bass
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No, that is the correct answer. 11.38 m/s to be "precise," but you're only given two significant figures in your initial conditions. I've done it three ways, and it comes to 11.38 m/s each time. And 40 m/s is correct for a. Just now, I did it a 4th way, and I got 10.96 m/s - which still rounds to 11 m/s.
 
  • #6
Ohoneo
22
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No, that is the correct answer. 11.38 m/s to be "precise," but you're only given two significant figures in your initial conditions. I've done it three ways, and it comes to 11.38 m/s each time. And 40 m/s is correct for a. Just now, I did it a 4th way, and I got 10.96 m/s - which still rounds to 11 m/s.

Okay, so I did do the math correctly. Thanks for your help :) I'm going to email my professor and see what the issue is with that problem.
 

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