Dynamics - supertanker coasting to a stop

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The discussion centers on calculating the distance a supertanker travels while coasting to a stop from an initial speed of 30 km/h over 21 minutes, assuming constant negative acceleration. The correct distance traveled is determined to be 5.25 km, not the initially calculated 315 km. The force exerted by fluid friction while stopping is calculated to be approximately 1.65 million Newtons. Participants clarify the importance of using the correct mass and speed values in the calculations. The thread highlights the significance of careful unit conversion 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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