Question

Find the acceleration in g:

A test of the prototype of a new automobile
shows that the minimum distance for a controlled stop from 99 km/h to zero is 59 m.
Find the acceleration, assuming it to be
constant as a fraction of the free-fall acceleration. The acceleration due to gravity is
9.81 m/s^2.
Answer in units of g.

Answer 1

I've been trying to solve this for an hour now, and I know I'm using the right equation, I just feel like the units are probably messing me up.

I've used v^2=vi^2 + 2ay, and plugged this in 0^2=99^2+2a(.059). I changed 59m to .059 kilometers so it'd be the same as 99 km/h, but is that the right conversion even? Even then I get -83059.32203 km/h^2, then convert it to m/s^2 and get -23072.0339 m/s^2. Then I divide it by 9.81 to get it into g's and get -2351.889 g but that is wrong. What am I doing wrong?

Answer 2

When doing kinematics problems I prefer converting to m/s first:\[99km/hr = 27.5m/s\]Using the equation you get:\[V_f^2=V_i^2+2ad\]\[(0m/s)^2=(27.5m/s)^2+2a(59m)\]\[a = -6.4089 m/s^2\]
Convert that to a fraction of g:\[\frac{-6.41m/s^2}{-9.81m/s^2}=\text{~}0.65g\]

Answer 3

That seems correct, and I get your thinking but it says the answer is wrong. Although, the second part of the question I have to solve for how long it'll take the car to stop in seconds, and I used the acceleration of -6.4 and got the correct time amount. It may just be the question though, I'm not sure.

Thank you so much! Those conversions always get me.

Answer 4

It may just be a difference in significant digits but I'm glad I could help :)