## anonymous 4 years ago A car slows down from 20 m/s to rest in a 85m distance. What was its acceleration if assumed constant?

1. anonymous

So far I've set up the problem as such: the initial velocity is 20 m/s, the final velocity is 0m/s, the distance is 85m and I want the acceleration.

2. JamesJ

By the Work-Energy theorem, the change in kinetic energy must be equal to the work done: $\Delta KE = Work$ Now $$Work = Fd = mad$$. You should now have enough to solve for acceleration $$a$$.

3. anonymous

I haven't learned that yet. This is using the kinematic equations. I was thinking of using either vf^2=vi^2+2ad where vf is final velocity, vi is initail velocity, a is acceleration and d is distance or d=vi + vf/2 but when I plugged in my values for weach equation, I got 570 and 8.5, which were wrong answers.

4. JamesJ

This is where that equation comes from. $\Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$ Now by the Work-Energy theorem, the change in energy of a system is equal to the work done on it, $\Delta KE = Fs = mas$ where $$s$$ is displacement. Hence we have that $\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mas$ Cancelling now $$m$$ from both sides and multiplying through by 2, we have $v_f^2 - v_i^2 = 2as$

5. JamesJ

Or in other words, $v_f^2 = v_i^2 + 2ad$ writing now $$d$$ for displacement. In your problem, you know everything in this equation except acceleration $$a$$.

6. JamesJ

Because $$v_f = 0$$, you have $a = -\frac{v_i^2}{2d}$

7. anonymous

i see.

8. anonymous

If i have the right equation, then I must be evaluating the problem wrong. I just plugged in each variable as such: (0m/s)^2 = (20m/s)^2+2a(85m)

9. anonymous

and my final answer was 570m/s^2. Maybe I need to divide the 400 by 170....

10. JamesJ

Using the equation I wrote down, $a = -\frac{v^2}{2d} = -\frac{400}{170} = 2.35$

11. JamesJ

*dropped the negative $a = -2.35 \ m/s^2$

12. anonymous

yes I ended up with the same answer. I figured out what I did wrong thanks.