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Final velocity given acceleration and initial velocity 
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#1
Jan2511, 09:21 AM

P: 22

1. The problem statement, all variables and given/known data
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. 2. Relevant 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(xx0) 3. 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! 


#2
Jan2511, 09:53 AM

Mentor
P: 11,689

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
Jan2511, 09:57 AM

P: 137

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
Jan2511, 10:13 AM

P: 22

Final velocity given acceleration and initial velocity



#5
Jan2511, 10:19 AM

P: 137

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
Jan2511, 10:40 AM

P: 22




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