When a train is traveling along a straight track at 2 m/s, it begins to accelerate at (60 v^{-4}) m/s^2 where v is in m/s. Determine its velocity v and the position 3 s after the acceleration.

Image from: R. Hibbeler, Engineering Mechanics Dynamics, 13th ed. Upper Saddle River, N.J.: Pearson, 2013.
Solution:
Show me the final answer↓
Remember that acceleration is:
We will isolate for \text{d}t.
The question asks us to find the velocity after 3 seconds, thus, we will integrate the left side from 0 to 3 seconds. On the right side, we know the initial velocity to be 2 m/s, and so we will integrate from 2 to v where v is final velocity.
(Solve for v)
v=3.93 m/s
To figure out the position, we will use the following kinematic equation.
(isolate for \text{d}s)
\text{d}s=\dfrac{v\text{d}v}{a}Remember that:
\text{d}s=\dfrac{v\text{d}v}{a}=\dfrac{1}{60}v^5\text{d}v
Again, we will integrate both sides of the equation.
s=9.98 m
Final Answers:
s=9.98 m
Hi! I would like to ask the part that I don’t clearly get it. Where did 1/300 (v^5-32) came from?
So you’re just solving the left side of the integral. If you solve it, that’s the value you get.
hi i would like to ask why did you use 3.92 and 2 for the last equation?
So the initial velocity is 2 m/s (given in the question), and the final velocity we found, which was 3.93 m/s. Using that, we can figure out the distance.
I feel lost in the integration part, firstly the integration of dt is t and when we replaced we got 3 is that right? Then when we integrated the right side how did it come as 1/300(v^5-32)? Can you please explain in a little more details
Thank you
Never mind I got it 🙂
Glad to hear!
How did you compute the last part? How did you come in a 9.98m?
Make sure to solve the integral properly.
For the distance part firstly why did you use 1/60v^5 alone without the 1/5 because it was 1/300, also why you didn’t use the the equation v=ds/dt instead?
Thank you
It’s the value you get after you take the integral. And it’s not possible to use v=ds/dt, we don’t have enough givens to use that equation.
Please I need more solved problems on kinematics so I can prepare for my test in few days to come
i want to ask, how did you get the final answer for s = 9.98m but when i calculated it, i get 10.07m
Did you solve the integral properly? Might be a numerical error.