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.

#### Solution:

Show me the final answerâ†“

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

#### Final Answers:

v=3.93 m/s

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 đź™‚

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 didnâ€™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.