Question

calculate the magnitude of a bullets acceleration if it travels 1200 m/s and stops within a bulletproof vest that is 1.0cm thick

Answer 1

\[V _{fin}=V _{initial} + 2*a*x \]

You can use this.

note that distance is in meter unit. So you should change cm into meter.

Answer 2

You'd be better off using the equation
\[v_{f}^{2} = v_{i}^{2}+2a(x_{f}-x_{i})\]

Answer 3

upps sorry. I have forgotten the square terms.

Answer 4

proof:
acceleration (a) is a change of velocity over time
where the change in velocity is initial velocity v0 minus final velocity v\[a=\frac{v-v_{0}}{t}\]solving for t we get\[t=\frac{v-v_{0}}{a}\]
For constant acceleration the average acceleration is midway between v0 and v\[a_{av1}=\frac{v_{0}+v}{2}\]average acceleration is also the change in x over time\[a_{av2}=\frac{x-x_{0}}{t}\]\[a_{av1}=a_{av2}\]\[\frac{v_{0}+v}{2}=\frac{x-x_{0}}{t}\]solve for x\[x=x_{0}+(\frac{v_{0}+v}{2})t\]substitute t\[x=x_{0}+(\frac{v_{0}+v}{2})(\frac{v-v_{0}}{a})\]\[x=x_{0}+\frac{v_{0}v-v_{0}^{2}+v^{2}-v_{0}v}{2a}\]\[x=x_{0}+\frac{v^{2}-v_{0}^{2}}{2a}\]\[v^{2}=v_{0}^{2}+2a(x-x_{0})\]

Answer 5

oh thank you so much :)