Question

two kittens are on opposite sides of a field, 250 m apart. kitten the runs at a constant speed of 25 m/s due east on a collision course with kitten b, which is traveling west at 12 m/s. how much time elapses before the two kittens collide?

Answer 1

Let the starting position of the kitten on the left side (call it A) of the field (running east, which will serve as the positive direction) act as the origin. The other kitten (call it B) then has a starting position of ${x_B}_0=250\,\mathrm m$ while A has a starting position of ${x_A}_0=0\,\mathrm m$.

A is traveling at a velocity of $v_A=25\,\dfrac{\mathrm m}{\mathrm s}$, while B is traveling at a velocity of $v_B=-12\,\dfrac{\mathrm m}{\mathrm s}$. Their respective positions over time are given by

$x_A=\left(25\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$x_B=250\,\mathrm m+\left(-12\,\dfrac{\mathrm m}{\mathrm s}\right)t$

A and B will collide at the point when $x_A=x_B$, so we solve:

$\left(25\,\dfrac{\mathrm m}{\mathrm s}\right)t=250\,\mathrm m+\left(-12\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$\implies\left(37\,\dfrac{\mathrm m}{\mathrm s}\right)t=250\,\mathrm m$

$\implies t=\dfrac{250\,\mathrm m}{37\,\frac{\mathrm m}{\mathrm s}}=6.76\,\mathrm s$

or about 6.8 s if taking significant digits into account.

Answer 2

The time elapse before the two kittens collide will be "6.76 s".

According to the question,

The dividing distance,

• 250 m

Resultant velocity,

= 25 m/s + 12 m/s

= 37 m/s

hence,

The time will be:

→ $Time = \frac{Distance}{Speed}$

By substituting the values,

            $= \frac{250}{37}$

            $= 6.76 \ s$

Thus the approach above is right.

Learn more about elapse here:

https://brainly.com/question/1931458