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Assume a car has hit a wall in a right angled collision and the front bumper has been displaced 9 cm. The resulting impact is 25g. Also, it is evident by skid marks that the car braked for 5m with an acceleration of 1.5m/s^2. What is the impact velocity in this collision?

Here's what I get out of it. $$\begin{align}\Delta d &=0.09\text{ m}\\ a &= 196\ \mathrm{m/s^2}\\ V_2 &= 0\text{ m/s}\\\end{align}$$

Then I determine $V_1$ by:

$$\begin{align}V_2^2 &= V_1^2 + 2a\Delta d\\ 0 &= V_1^2 -35.28 \\ 5.94\text{ m/s} &= V_1\end{align}$$

My textbook does not give this answer. Could anyone please explain why. I have been looking at it for hours.

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Also, I realized that my formatting did not work. I am sorry. –  John Feb 17 '12 at 1:24
    
No problem, I fixed that up for you. Do you have any ideas about things you might be doing wrong? Anything you're not sure about? Does the value of the answer given in the textbook suggest anything you might be doing wrong? If you have any insights of this sort, it would be good stuff to include in your question. –  David Z Feb 17 '12 at 1:33
    
I have a feeling that my textbook is wrong. But, I just wanted to double check. It is very likely as the "textbook" was created by my teacher because he is too cheap to purchase real published books for the school. –  John Feb 17 '12 at 1:34
    
OK, gotcha. You could (but you don't have to) mention in your question that you suspect the textbook answer might be wrong, and why. It's good to know when there is a good chance that the book is wrong, because with popular textbooks, that is exceedingly rare. –  David Z Feb 17 '12 at 1:40
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2 Answers 2

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I always look for the easiest way to describe questions like this, and I think the easiest way is to do it in reverse. Start with the car stationary and accelerate it at 25g for 9cm, then accelerate it at 1.5ms$^{-1}$ for 5m. You've correctly identified the relevant equation of motion:

$$v^2 = u^2 + 2as$$

So for step 1 i.e. the bumper deforming:

$$v_1^2 = 2 \times 25g \times 0.09$$

Then for step 2 i.e. the braking:

$$v_2^2 = v_1^2 + 2 \times 1.5 \times 5 = 2 \times 25g \times 0.09 + 2 \times 1.5 \times 5$$

See if that gives you the same answer as your book.

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Why is $a=196 m/s^2$? It's 20g, not 25g. And it looks like you did not take into account the deceleration before impact.

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answers on stack exchange are meant to be more than just correcting numbers... –  Nic Feb 17 '12 at 11:49
    
I have been criticized here for too complete answers to homework questions, you scold me for an incomplete answer:-) –  akhmeteli Feb 17 '12 at 15:10
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