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.

When a train is traveling along a straight track

Image from: R. Hibbeler, Engineering Mechanics Dynamics, 13th ed. Upper Saddle River, N.J.: Pearson, 2013.

Solution:

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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

 

This question can be found in Engineering Mechanics: Dynamics, 13th edition, chapter 12, question 12-2.

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