Deceleration question: When a cyclist reaches the bottom of a large hill, he is travelling at v = 8m/sec

[bluefrog]

I am attempting this problem, which is on pg 101 of of the Pearson Year 1 AS student book.

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When a cyclist reaches the bottom of a large hill, he is travelling with a speed, [imath]\footnotesize{v}[/imath], of [imath]\footnotesize{8\, \textrm{ms}^{-1}}[/imath]. The cyclist then free-wheels along a straight flat road with speed [imath]\footnotesize{v = (8 - \sqrt{t\;})\,\textrm{ms}^{-1}}[/imath], where [imath]\footnotesize{t}[/imath] is the time elapsed since he reached the bottom of the hill.

Find:

a the time it takes for the cyclist to come to a stop

b [imath]\footnotesize{\dfrac{dv}{dt}}[/imath]

c the cyclist's acceleration at time [imath]\footnotesize{t = 9}[/imath] seconds.

Hint: Acceleration is the rate of change of speed with respect to time:

[imath]\footnotesize{\qquad a = \dfrac{dv}{dt}}[/imath]

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I am unsure how to solve a), I assumed differentiate and then equate to zero would give me the correct answer, which is 64 seconds (no working out is shown, only the answer is given in the back of the book), but t simply ends up as zero as well, which cannot be.

[imath]\qquad \text{given } v = 8 - \sqrt{t}, \text{ then } \frac{dv}{dt} = -\frac{1}{2}t^{-\frac{1}{2}}, \text{ which, when equated to zero result in } t= 0[/imath]

[imath]\qquad \text{When equated to } 8, \text{ then } -\frac{1}{2}t^{-\frac{1}{2}}=8 \Rightarrow t=\frac{1}{256}[/imath]

Any suggestions where I am going wrong would be appreciated.

 

[The Highlander]

I am unsure how to solve a), I assumed differentiate and then equate to zero would give me the correct answer, which is 64 seconds (no working out is shown, only the answer is given in the back of the book), but t simply ends up as zero as well, which cannot be.
Any suggestions where I am going wrong would be appreciated.

You're overcomplicating it!

Simply replace \(\displaystyle v\) with zero in the equation you are given (for his "speed" at any time, \(\displaystyle t\))...

[math]v= 8-\sqrt{t}[/math]
So, if [math]0=8-\sqrt{t}[/math]
What does \(\displaystyle t\) = ?

 

[bluefrog]

Ahhh, thanks. Got confused because of the hint given, which is for b) and c), but not a)

 

[The Highlander]

Ahhh, thanks. Got confused because of the hint given, which is for b) and c), but not a)

?

 

[Steven G]

I don't like this problem as the answer can be anything >64.
The reason is the problem does not state how long it takes the cyclist to stop once reaching the bottom of the hill. In theory, it could have taken the cyclist 4 hrs and 13 minutes to reach the bottom of the hill. Then, from the top of the hill it would have taken the cyclist 5 hrs and 9 minutes to stop.

Having said that, one can only assume that the author of the problem meant to ask how long does the cyclist take to stop once reading the bottom of the hill.

As a student I would always say that the problem couldn't be answered because the question was flawed. This way I would never make a silly mistake doing the problem the way I thought my professor meant and I would always get full credit.

I always felt that it was insulting to my professor for me to re-write their problem.

 

[Steven G]

After reading part b, I have the same comment. dv/dt is different for the cyclist going down the hill compared to dv/dt after reaching the bottom of the hill.

Part c: When is t=0???? It doesn't say anywhere in the problem that the clock starts once the cyclist reaches the bottom of the hill.

As far as an exam problem goes, I would love this problem as I feel no parts can be answered without making some (unfair) assumptions and I would get full credit for stating so.

 

[The Highlander]

After reading part b, I have the same comment. dv/dt is different for the cyclist going down the hill compared to dv/dt after reaching the bottom of the hill.

Part c: When is t=0???? It doesn't say anywhere in the problem that the clock starts once the cyclist reaches the bottom of the hill.

As far as an exam problem goes, I would love this problem as I feel no parts can be answered without making some (unfair) assumptions and I would get full credit for stating so.

Oh No! We're back to disagreeing again, Steven! ?
I'm afraid you haven't read the question carefully. ?
It quite clearly states: "\(\displaystyle t\) is the time elapsed since he reached the bottom of the hill."
Here is a nice little diagram of the situation described in the question (so you can understand it better ?).

        ?‍                                             ←?‍

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           |←⸺⸺⸺⸺⸺ 64 s ⸺⸺⸺⸺⸺→|
       \(\displaystyle v\) = 0 ms^-1                                        \(\displaystyle v\) = 8 ms^-1
       \(\displaystyle t\) = 64s                                            \(\displaystyle t\) = 0s
                             when \(\displaystyle v=(8-\sqrt{t})\)

Whatever might have happened on the hill is completely irrelevant, we are only interested in what happens along the the straight and level road starting off from the bottom of the hill at which point the clock starts with \(\displaystyle t\) equal to zero.

 

[Steven G]

Oh No! We're back to disagreeing again, Steven! ?
I'm afraid you haven't read the question carefully. ?
It quite clearly states: "\(\displaystyle t\) is the time elapsed since he reached the bottom of the hill."
Here is a nice little diagram of the situation described in the question (so you can understand it better ?).

        ?‍                                             ←?‍

---

           |←⸺⸺⸺⸺⸺ 64 s ⸺⸺⸺⸺⸺→|
       \(\displaystyle v\) = 0 ms^-1                                        \(\displaystyle v\) = 8 ms^-1
       \(\displaystyle t\) = 64s                                            \(\displaystyle t\) = 0s
                             when \(\displaystyle v=(8-\sqrt{t})\)

Whatever might have happened on the hill is completely irrelevant, we are only interested in what happens along the the straight and level road starting off from the bottom of the hill at which point the clock starts with \(\displaystyle t\) equal to zero.

There is nothing to disagree about other than I might need eye glasses. It clearly stated that t is the time elapsed since he reached the bottom of the hill.

 

[khansaheb]

When a cyclist reaches the bottom of a large hill, he is travelling with a speed, v\footnotesize{v}v, of 8ms−1\footnotesize{8\, \textrm{ms}^{-1}}8ms−1.

and

The cyclist then free-wheels along a straight flat road with speed v= (8 - √t) m * s^-1

shows that at t= 0, v = 8 m*s^-1. At the bottom of the hill the cyclist travelling with speed v = 8 m*s^-1 - the only time that speed is attained (at t=0).