Question

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

Answer 1

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.

Answer 2

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 \( a \).

Answer 3

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.

Answer 4

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
\[ v_f^2 - v_i^2 = 2as \]

Answer 5

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 \( a \).

Answer 6

Because \( v_f = 0 \), you have
\[ a = -\frac{v_i^2}{2d} \]

Answer 7

i see.

Answer 8

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)

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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