Can someone help? I am confuse why we can’t use the constant acceleration formula, v^2=v0^2+ 2a(delS)… it may be that this is NOT constant a, if so what makes it not constant? (The v var being multiplied?) sorry I have 15 years out of school and I am very rusty.
For it to be constant acceleration, the acceleration given would have to be a single numerical value without any other variables. Constant acceleration equation would be "a = 5", but not a linear equation like "a = 2v + 1" or an exponent function like the one given in the problem.
Hi Faroq, I get what you mean in the video, but how do you know a (the deceleration) is independent of the distance s? This should be the premise for the integration.
Hi! The given acceleration (a) is explicitly a function of the velocity (v). This alone is enough to solve the problem. Using the Particle Rectilinear Motion equation from the FE Handbook on page 102, we can express acceleration in terms of velocity (v) and distance (s) as: a * ds = v * dv Rearranging to isolate the acceleration term: a = v * (ds/dv) We can then substitute the given expression for acceleration, resulting in: -1.5 * v^(1/2) = v * (ds/dv) Isolate "ds" and solve, then integrate with respect to velocity to find the total distance (s).
I believe in the solution we integrated "dS" not "S". "dS" is the "differential of S". Since integration is the opposite of differenitation, the integral of "dS" simply becomes S. I hope that makes sense.
Technically, it’s supposed to be there but for almost all dynamics problems we ignore it because it will be ZERO. This is because when the initial velocity V = 6 m/s, the distance S = 0, giving us a constant of integration, C = 0 snipboard.io/RdcFEk.jpg
@@directhubfeexam Thank you so much for responding! So when s=0 and v=6, C = 0.444(6)^3/2-6.532=-0.0066. Is the only reason it's not zero due to rounding errors?
Very clear explanation and make it really easy. Thanks a lot!
Can someone help? I am confuse why we can’t use the constant acceleration formula, v^2=v0^2+ 2a(delS)… it may be that this is NOT constant a, if so what makes it not constant? (The v var being multiplied?) sorry I have 15 years out of school and I am very rusty.
I was wondering the same thing
For it to be constant acceleration, the acceleration given would have to be a single numerical value without any other variables. Constant acceleration equation would be "a = 5", but not a linear equation like "a = 2v + 1" or an exponent function like the one given in the problem.
Hi Faroq, I get what you mean in the video, but how do you know a (the deceleration) is independent of the distance s? This should be the premise for the integration.
Hi!
The given acceleration (a) is explicitly a function of the velocity (v). This alone is enough to solve the problem.
Using the Particle Rectilinear Motion equation from the FE Handbook on page 102, we can express acceleration in terms of velocity (v) and distance (s) as:
a * ds = v * dv
Rearranging to isolate the acceleration term:
a = v * (ds/dv)
We can then substitute the given expression for acceleration, resulting in:
-1.5 * v^(1/2) = v * (ds/dv)
Isolate "ds" and solve, then integrate with respect to velocity to find the total distance (s).
@@directhubfeexam Thank you, this is a great explanation. I'll be taking my fe next week, I'm sure that I'll pass it easily with you help.
@@xzxz1506 You're very welcome 😁
Wishing you all the best this week! Don’t forget to take a break the day before the exam. Go in there and crush it!
When we integrate the "S", why didn't it become S^2/2?
I believe in the solution we integrated "dS" not "S". "dS" is the "differential of S". Since integration is the opposite of differenitation, the integral of "dS" simply becomes S.
I hope that makes sense.
@@directhubfeexam that does make sense. Thank you
@@zaq1525 You're welcome!
Awesome videos, Thank you
Why was there no +C after we integrated s at the end?
Technically, it’s supposed to be there but for almost all dynamics problems we ignore it because it will be ZERO. This is because when the initial velocity V = 6 m/s, the distance S = 0, giving us a constant of integration, C = 0
snipboard.io/RdcFEk.jpg
@@directhubfeexam Thank you so much for responding!
So when s=0 and v=6, C = 0.444(6)^3/2-6.532=-0.0066. Is the only reason it's not zero due to rounding errors?
@@tjweilerexactly! It’s a rounding error. I’m glad we cleared this up.
@directhubfeexam thanks for responding! I'm preparing for the FE, and your videos have been really helpful! You're awesome!
@@tjweiler Thank you very much! I'm really happy to hear this.
Thank you.