Friction and distance traveled

[osker246]

Homework Statement

Sam, whose mass is 67 kg, takes off across level snow on his jet-powered skis. The skis have a thrust of 230 N and a coefficient of kinetic friction on snow of 0.13. Unfortunately, the skis run out of fuel after only 12.

Homework Equations

F-Fk=mA

(Vx)f = (Vx)i + (Ax)(Delta Time)
Xf=Xi+((Vx)i)(delta Time) + .5(Ax)(Delta Time)^2

The Attempt at a Solution

Well I first started out by find the acceleration:

230-(.13)(67)(9.8)=67Ax

Ax=2.15

So I plugged that into the equation to find the max speed...

(Vx)f= 2.15 x 12 = 25.8 m/s rounded to 2 sig figures 26 m/s

According to mastering physics I am correct up to here. Finding the distance traveled is where I am getting the answer incorrect.

I tried using the equation Xf=Xi+((Vx)i)(delta Time) + .5(Ax)(Delta Time)^2

Xf= 0 + (26)(12) + .5(2.15)(12)^2 = 466.8 M

Mastering physics ask for it to be rounded to 2 sig figures giving 470 m. But yet that answer is incorrect. What am I doing wrong? If somebody can help me I'd appreciate it. Thanks!

 

[Doc Al]

Are you asked to find the total distance the skier travels up to the point where he comes to rest, or just the distance he travels until the jet-powered skis run out of fuel? (Once the jets stop, what is the skier's acceleration?)

(Get in the habit of posting the complete problem exactly as given.)

 

[osker246]

I'm sorry that's my fault. I forgot to type the questions asked.

Part A:"What is Sam's top speed?"

I got this part correct. Part B is where I am lost.

Part B:"How far has Sam traveled when he finally coasts to a stop?"

 

[Doc Al]

Part B:"How far has Sam traveled when he finally coasts to a stop?"

Just as I thought.

You'll need to solve this problem in two parts, since each part has a different acceleration. First solve for the distance traveled while he's speeding up, using the formula for distance you already used. Then solve for the distance traveled while he's slowing down. (First figure out his new acceleration.) You'll need to use a different kinematic formula, since you don't have the time.

Add those two distances to get the total.