Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am trying to solve the following:

A man of mass 83 kg jumps down to a concrete patio from a window ledge only 0.48 m above the ground. He neglects to bend his knees on landing, so that his motion is arrested in a distance of about 2.2 cm. What is the average acceleration of the man from the time his feet first touch the patio to the time he is brought fully to rest?

Now, I know that the man has a velocity of $v_1=\sqrt{2gy}$, where $y_1=0.48$ m, when his feet first touch the patio. From the equation of motion for constant acceleration, $a=\frac{v_1^2}{2y_2}$, where $y_2=2.2$ cm. However, I am not sure if I can use this since the equation of motion is derived for constant acceleration. What do I do in the case of average acceleration? Do I need an additional assumption such as constant jerk?

share|improve this question

2 Answers 2

up vote 0 down vote accepted

Take an example: Suppose a particle constrained to move on the $x$-axis is at the origin. Now suppose after $10$ seconds it is $30$ meters away from the origin. We can then say that the particle's average velocity was $3$ m/s. But of course, in reality it may have done all sorts of detouring whilst changing its velocity such that after $10$ s it ends up $30$ m away from the origin. What the $3$ m/s tells us is that had it been moving with a constant velocity during this time interval, it would be $3$ m/s because only then it could have covered $30$ m in $10$ seconds. Similarly, if the particle's velocity is first $0$ and then $50$ m/s after $5$ seconds, we'd say that its average acceleration was $10$ m/s^2 which means that had it been moving with a constant acceleration, it would be equal to this. Average acceleration simply gives us a number which when multiplied by the time interval gives us the final velocity. This number is exactly equal to the acceleration it would have had if it was constant throughout the object's motion.

Also, if this helps: Average acceleration is defined as $a_{avg} = (v_f - v_i)/(t_f-t_i)$. If you multiply both sides by $(t_f-t_i)$ you immediately get one of the equations of motion. Then adding $v_i$ to both sides gives: $a_{avg}(t_f-t_i)+ v_i = v_f =\frac{dv}{dt}$. You can now integrate both sides to get another equation of motion. Notice that $a_{avg}$ is just a number so you can take this out of the integrand. And then of course you can eliminate the time variable by substituting one of the equations into the other to get the third equation of motion which you are using in your problem.

share|improve this answer

HINT : Assume constant acceleration

share|improve this answer
    
My question is asking why I can assume constant acceleration. –  jay Mar 3 '13 at 0:15
    
I will be honest with you. Because you have no other choice. –  Cheeku Mar 3 '13 at 3:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.