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Jason takes off from rest across level water on his jet-powered skis. The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible). The skis have a thrust of 200 N and a coefficient of kinetic friction on water of 0.10. Unfortunately, the skis run out of fuel after only 11 s. How far has Jason traveled when he finally coasts to a stop

Answer :

Answer:

The total distance traveled by body is = 274.41 m

Explanation:

Given :

Combined mass [tex]m = 75[/tex] Kg

Thrust force [tex]F_{t} =[/tex] 200 N

Coefficient of kinetic friction [tex]\mu _{k} =[/tex] 0.10

According to newton's law

 [tex]F_{t} - \mu m g = ma[/tex]

[tex]200 - 0.1 \times 75 \times 9.8 = 75 a[/tex]    ( ∵ [tex]g = 9.8 \frac{m}{s^{2} }[/tex] )

Before run out of fuel the acceleration is,

[tex]a = \frac{126.43}{75} = 1.69\frac{m}{s^{2} }[/tex]

Distance covered in 11 sec. is given by,

  [tex]y = v_{0}t + \frac{1}{2}at^{2}[/tex]

Where [tex]y =[/tex] distance, [tex]v_{0} =[/tex] initial velocity = [tex]0[/tex], [tex]a =[/tex] acceleration,

  [tex]y = \frac{1}{2} \times 1.69 \times (11) ^{2}[/tex]

  [tex]y = 102.24[/tex] m

After run out of fuel velocity of body is,

[tex]v = v_{o} + at[/tex]

Where [tex]v_{o} =[/tex] 0

[tex]v = 1.69 \times 11 = 18.37 \frac{m}{s}[/tex]

After fuel run out of only frictional force act on body,

[tex]ma = \mu_{k} m g[/tex]

Acceleration is given by,

 [tex]a = 0.1 \times 9.8[/tex]

 [tex]a = 0.98 \frac{m}{s^{2} }[/tex]

So distance traveled by body after run out of fuel is given by

   [tex]v^{2} - v_{o} ^{2} =2ax[/tex]

Where  initial velocity [tex]v_{o} =[/tex] 0,

   [tex]x = \frac{337.46}{1.96}[/tex]

   [tex]x = 172.17[/tex] m

So total distance traveled by body is,

[tex]d = 172.17 +102.24 = 274.41[/tex] m

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