# Thread: Finding total time coasting to shore?

1. ## Finding total time coasting to shore?

As a prank, your friends have kidnapped you in your sleep, and transported you out onto the ice covering a local pond. When you wake up you are 27.0 m from the nearest shore. The ice is so slippery (i.e. frictionless) that you cannot seem to get yourself moving. You realize that you can use Newton's third law to your advantage, and choose to throw the heaviest thing you have, one boot, in order to get yourself moving. Take your weight to be 565 N.

If you throw your 1.23-kg boot with an average force of 386 N, and the throw takes 0.640 s (the time interval over which you apply the force), what is the magnitude of the force that the boot exerts on you? (Assume constant acceleration.)
to this I got 386

However, I'm having a hard time finding this answer: How long does it take you to reach shore, including the short time in which you were throwing the boot?

I tried saying that total time = time throwing + time coasting
and I used the equation distance to shore=0.5a(t^2) + at(throw)t(coast)
the answer I got was 0.8256, but it is the wrong answer. Any help?

2. Originally Posted by horan6
As a prank, your friends have kidnapped you in your sleep, and transported you out onto the ice covering a local pond. When you wake up you are 27.0 m from the nearest shore. The ice is so slippery (i.e. frictionless) that you cannot seem to get yourself moving. You realize that you can use Newton's third law to your advantage, and choose to throw the heaviest thing you have, one boot, in order to get yourself moving. Take your weight to be 565 N.

If you throw your 1.23-kg boot with an average force of 386 N, and the throw takes 0.640 s (the time interval over which you apply the force), what is the magnitude of the force that the boot exerts on you? (Assume constant acceleration.)
to this I got 386

However, I'm having a hard time finding this answer: How long does it take you to reach shore, including the short time in which you were throwing the boot?

I tried saying that total time = time throwing + time coasting
and I used the equation distance to shore=0.5a(t^2) + at(throw)t(coast)
the answer I got was 0.8256, but it is the wrong answer. Any help?
$M = \frac{565}{g}$

$F \Delta t = M \Delta v$

$\frac{F \Delta t}{M} = v_f - v_0$

since $v_0 = 0$ ...

$\frac{386(0.64)}{M} = v_f
$

acceleration during the impulse period ... $a = \frac{v_f-v_0}{\Delta t}$

$t_{total} = 0.64 + \frac{27 - \frac{1}{2}a(0.64)^2}{v_f}$