# When a train is traveling along a straight track 14

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:

Remember that acceleration is:

$a=\dfrac{\text{d}v}{\text{d}t}$

We will isolate for $\text{d}t$.

$\text{d}t=\dfrac{\text{d}v}{a}$

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.

$\int_{0}^{3}\text{d}t=\int_{2}^{v} \dfrac{\text{d}v}{60v^{-4}}$

$3=\dfrac{1}{300}(v^5-32)$

(Solve for v)

$v=3.93$ m/s

To figure out the position, we will use the following kinematic equation.

$a\text{d}s=v\text{d}v$

(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.

$\int_{0}^{s}\text{d}s=\dfrac{1}{60}\int_{2}^{3.93}v^5\,\text{d}v$

$s=\dfrac{1}{60}\left(\dfrac{v^6}{6}\right)\Big|_{2}^{3.93}$

$s=9.98$ m

$v=3.93$ m/s

$s=9.98$ m

## 14 thoughts on “When a train is traveling along a straight track”

• Sheena

Hi! I would like to ask the part that I don’t clearly get it. Where did 1/300 (v^5-32) came from?

• questionsolutions Post author

So you’re just solving the left side of the integral. If you solve it, that’s the value you get.

• bobby

hi i would like to ask why did you use 3.92 and 2 for the last equation?

• questionsolutions Post author

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.

• chloe

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

• Chloe

Never mind I got it 🙂

• questionsolutions Post author

• How did you compute the last part? How did you come in a 9.98m?

• questionsolutions Post author

Make sure to solve the integral properly.

• Chloe

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

• questionsolutions Post author

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.

• Abubakar Abdulsalam

Please I need more solved problems on kinematics so I can prepare for my test in few days to come

• wan

i want to ask, how did you get the final answer for s = 9.98m but when i calculated it, i get 10.07m

• questionsolutions Post author

Did you solve the integral properly? Might be a numerical error.