MHB How do I derive and solve the equation for the athlete's running speed?

[Simonio]

I'm having problems getting going on the following question, any help appreciated:

As part of his training an athlete usually runs 80 km at a steady speed of \(v\) km h. One day he decided to reduce his speed by 2.5 km h and his run takes him an extra 2h 40 mins.

Derive the equation \(\frac{80}{v}+ \frac{8}{3} = \frac{160}{2v-5}\) and solve it to find the value of \(v\).

 

[kaliprasad]

I'm having problems getting going on the following question, any help appreciated:

As part of his training an athlete usually runs 80 km at a steady speed of \(v\) km h. One day he decided to reduce his speed by 2.5 km h and his run takes him an extra 2h 40 mins.

Derive the equation \(\frac{80}{v}+ \frac{8}{3} = \frac{160}{2v-5}\) and solve it to find the value of \(v\).

where is the problem deriving or solving

 

[I like Serena]

I'm having problems getting going on the following question, any help appreciated:

As part of his training an athlete usually runs 80 km at a steady speed of \(v\) km h. One day he decided to reduce his speed by 2.5 km h and his run takes him an extra 2h 40 mins.

Derive the equation \(\frac{80}{v}+ \frac{8}{3} = \frac{160}{2v-5}\) and solve it to find the value of \(v\).

Hi Simonio!

Suppose we define $t$ to be the time it takes the athlete to run 80 km at a speed of $v$ km/h.
Can you set up the equations that correspond to the run times?

 

[Simonio]

Hi Simonio!

Suppose we define $t$ to be the time it takes the athlete to run 80 km at a speed of $v$ km/h.
Can you set up the equations that correspond to the run times?

Well I think I can say that \(t\)= \(\frac{80}{v}\) and with the slower speed \(t\) = \(\frac{80}{v-2.5}\) not sure of next step

 

[I like Serena]

Well I think I can say that \(t\)= \(\frac{80}{v}\)

Good! :)

and with the slower speed \(t\) = \(\frac{80}{v-2.5}\)

With the slower speed it takes 2h40 longer, which is $2\frac 2 3$ hours.
So it should be:
$$t + 2\frac 2 3 = \frac{80}{v-2.5}$$

not sure of next step

Can you combine the 2 equations and eliminate $t$?

 

[Simonio]

Good! :)
With the slower speed it takes 2h40 longer, which is $2\frac 2 3$ hours.
So it should be:
$$t + 2\frac 2 3 = \frac{80}{v-2.5}$$
Can you combine the 2 equations and eliminate $t$?

OK-you've given me a good hint there!

so: \(\frac{80}{v} + \frac{8}{3} = \frac{160}{2v-5}\)

 

[Simonio]

OK-you've given me a good hint there!

so: \(\frac{80}{v} + \frac{8}{3} = \frac{160}{2v-5}\)

Not sure about the next bit: do I multiply throughout by \(2v-5\)?

That would make: \((2v-5) \frac{80}{v} + (2v-5) \frac{8}{3} = 160\)

Think I'm on the wrong track

 

[I like Serena]

OK-you've given me a good hint there!

so: \(\frac{80}{v} + \frac{8}{3} = \frac{160}{2v-5}\)

Good!

Not sure about the next bit: do I multiply throughout by \(2v-5\)?

That would make: \((2v-5) \frac{80}{v} + (2v-5) \frac{8}{3} = 160\)

Think I'm on the wrong track

You're right on track.
Continue by multiplying throughout by $v$.

 

[Simonio]

Good!
You're right on track.
Continue by multiplying throughout by $v$.

Do you mean multiply by \(2v-5\)?

 

[I like Serena]

Do you mean multiply by \(2v-5\)?

You already did that.
But you're still left with $v$ in a denominator.

 

[Simonio]

You already did that.
But you're still left with $v$ in a denominator.

Then I get: \(v(2v-5)\frac{80}{v}\ +v(2v-5)\frac{8}{3} = 160v\)

= \(80(2v^2-5) + (2v^2-5)\frac{8}{3}\ = 160v\)

= \(160v^2 - 400v + \frac{16v^2-40v}{3}\ = 160v\)

Not sure whether I've lost the plot here!

 

[I like Serena]

Then I get: \(v(2v-5)\frac{80}{v}\ +v(2v-5)\frac{8}{3} = 160v\)

= \(80(2v^2-5) + (2v^2-5)\frac{8}{3}\ = 160v\)

= \(160v^2 - 400v + \frac{16v^2-40v}{3}\ = 160v\)

Not sure whether I've lost the plot here!

When you multiply $v(2v-5)$, you should get $(2v^2-5v)$ instead of $(2v^2-5)$.
That is because generally $a(b+c)=ab+ac$.
You can check this with for instance $4(2+3)$.

Furthermore, you can write:
$$v(2v-5)\frac{80}{v} = (2v-5)\frac{80}{v}v = (2v-5)80$$
That is because if you first divide by $v$ and then multiply by $v$ the effects cancel.