Dynamics - supertanker coasting to a stop

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SUMMARY

The discussion focuses on calculating the distance a supertanker, with a mass of 250,000 metric tons, travels while coasting to a stop from an initial speed of 30 km/h over a duration of 21 minutes. Using the formula S-S0 = (V+V0)t/2, the correct distance traveled is determined to be 5.25 km. Additionally, the force exerted by fluid friction during the stopping process is calculated to be approximately 1.65 x 10^6 N. The discussion highlights the importance of accurate unit conversions and formula application in physics problems.

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Dynamics -- supertanker coasting to a stop

1. Problem
Assuming constant negative acceleration, how far does a supertanker of a mass of 250000 metric tons travel while coasting to a stop from an initial speed of 30 km/h in 21 min?
What force is supplied by fluid friction of the water while the tanker is stopping?

2. Relevant formulas
S-S0 = (V+V0)t/2
ma = sum (F)
(1/2)at^2+v0t+S0=0
v=at+v0

3. Work I Did
250k met ton = 250,000,000 kg
30 km/h = 500 m/s
21 min = 1260 s
0 = 1260a + 500
a = -500/1260
(0+500)/2*1260 = 315 km (part a)
ma = -F(friction)
1000*(-50/126); Ff = 396.82 N (part b)

a: 5.25 km
b: 1.65*10^6 N
 
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30km/h is not 500 m/s !
 
Thanks!
Using the same equations as I did in part a I got 5250 m -> 5.25 km; a = -5/756; v0 = 50/6.
Part B I used 250,000,000 for the mass instead of 1000 kg and got the right answer.
This is kind of embarrassing. I must've stared at the problem, confused, for quite a while.
 

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