Question

a woman stands on the bank of a frozen lake with a dog by her side. she skims a bone across the ice at a speed of 3ms. the bone slows down with deceleration 0.4 ms square and the dog chases it with acce;eration 0.6 ms square. how far out from the bank does the dog catch up with the bone

Answer 1

\[x=3\frac{ m }{ s } \Delta t + \frac{ 1 }{ 2 }(-0.4\frac{ m }{ s^2 }) \Delta t^2\]

\[x= \frac{ 1 }{ 2 }(0.6\frac{ m }{ s^2 }) \Delta t^2\]

Answer 2

\[\frac{ 1 }{ 2 }(0.6\frac{ m }{ s^2 }) \Delta t^2=3\frac{ m }{ s } \Delta t + \frac{ 1 }{ 2 }(-0.4\frac{ m }{ s^2 }) \Delta t^2\]

Answer 3

if we multiply both sides by 2 we get
\[0.6\frac{ m }{ s^2 } \Delta t^2 = 6\frac{ m }{ s } \Delta t -0.4\frac{ m }{ s^2 } \Delta t^2\]

then we have to solve for t

Answer 4

from here we can actually just say
\[0.6t^2=6t-0.4t^2\]
divide t on both sides we get
\[0.6t=6-0.4t\]
solve for t
\[t=6\]

Answer 5

plug in 6 for t in both of these equations, to see if they are equal. and x=the distance when the dog catches up.

\[3\frac{ m }{ s } \Delta t + \frac{ 1 }{ 2 }(-0.4\frac{ m }{ s^2 }) \Delta t^2\]

\[ \frac{ 1 }{ 2 }(0.6\frac{ m }{ s^2 }) \Delta t^2\]