Final velocity given acceleration and initial velocity


by Ohoneo
Tags: acceleration, final, initial, velocity
Ohoneo
Ohoneo is offline
#1
Jan25-11, 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(x-x0)


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!
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gneill
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#2
Jan25-11, 09:53 AM
Mentor
P: 11,416
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.
p21bass
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#3
Jan25-11, 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.

Ohoneo
Ohoneo is offline
#4
Jan25-11, 10:13 AM
P: 22

Final velocity given acceleration and initial velocity


Quote Quote by p21bass View Post
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.
p21bass
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#5
Jan25-11, 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.
Ohoneo
Ohoneo is offline
#6
Jan25-11, 10:40 AM
P: 22
Quote Quote by p21bass View Post
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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