An oil tanker's engine have broken down and the wind has accelerated the tanker to a speed of 1.5m/s straight toward a reef. When the tanker is 500m from the reef, the wind dies down just as the engineer gets the engines going again. The rudder is stuck, so the only choice is to try to acclerate straight backwards away from the reef. The mass of the tanker and cargo is 3.6 X10^7 kg, and the engine produce a net horizontal force of 8.0X 10^4 N on the tanker. Will the ship hit the reef? If it does will the oil be safe? The hull can withstand an impact at a speed of .2 m/s or less.
I should have gotten .17m/s what is the equations I would use to get this answer?
Assume no friction, which for a ship in water is really a fantasy.
AverageForce*distance= 1/2 masscargoboat*v^2
solve for distance.
This problem is a far cry from reality.
sorry chris, physics stinks
To solve this problem, we need to use the principle of conservation of momentum. The initial momentum of the tanker is given by its mass multiplied by its initial velocity, which is 3.6 x 10^7 kg * 1.5 m/s = 5.4 x 10^7 kg·m/s.
When the wind dies down and the engine starts, the only force acting on the tanker is the horizontal force produced by the engine, which is 8.0 x 10^4 N. This force acts over a certain time period, so we need to find the change in momentum of the tanker during that time. We can use the equation F = Δp/Δt, where F is the force, Δp is the change in momentum, and Δt is the time.
Rearranging the equation, Δp = F * Δt.
Now that we have the equation for the change in momentum, we can find the time it takes for the tanker to stop. The change in momentum is equal and opposite to the initial momentum, so Δp = -5.4 x 10^7 kg·m/s.
Substituting the given values into the equation, we get -5.4 x 10^7 kg·m/s = (8.0 x 10^4 N) * Δt.
Solving for Δt, we find Δt = (-5.4 x 10^7 kg·m/s) / (8.0 x 10^4 N) = -675 s. Note that the negative sign indicates that the change in momentum is in the opposite direction to the initial momentum.
Now that we have the time it takes for the tanker to stop, we can calculate the distance it travels during that time using the equation d = v * t, where d is the distance, v is the velocity, and t is the time.
Substituting the given values into the equation, we get d = (1.5 m/s) * (-675 s) = -1012.5 m. Again, the negative sign indicates that the distance is in the opposite direction to the initial velocity.
Since the distance is negative, we can see that the tanker will not be able to stop before hitting the reef, as it only has 500 m of distance left. Therefore, the ship will hit the reef.
As for the speed of impact, since the distance traveled during the stopping time is 1012.5 m and the time is 675 s, we can calculate the speed using the equation v = d / t.
Substituting the given values into the equation, we get v = (-1012.5 m) / (675 s) = -1.50 m/s. Again, the negative sign indicates that the speed is in the opposite direction to the initial velocity.
The absolute value of the speed is 1.50 m/s, which is greater than the maximum impact speed of 0.2 m/s that the hull can withstand. Therefore, the ship will hit the reef and the oil may not be safe.