Newton's Law of Motion for a Straight Line Motion

[!!!]

Homework Statement

An oil tanker's engines have broken down, and the wind is blowing the tanker straight toward a reef at a constant speed of 1.5 m/s. When the tanker is 500 m 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 accelerate straight backward away from the reef. The mass of the tanker and cargo is 3.6 x 10⁷ kg, and the engines produce a net horizontal force of 8.0 x 10⁴N 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 0.2 m/s or less. You can ignore the retarding force of the water on the tanker's hull.

Homework Equations

F = ma
v_x = v_0x + a_xt
x = x₀ + v_0xt + 1/2 a_xt²
v_x²= v_0x² + 2a_x(x-x₀)
x - x₀ = (v_0x + v_x / 2)t

The Attempt at a Solution

I don't exactly know what to do first, so I first found the acceleration of the ship's engines.
a = f/m = 8.0 x 10⁴N / 3.6 x 10⁷ kg = 2.22 x 10³ m/s²

Then I tried to find the time it takes for the ship to hit the reef:
v_x = v_0x + a_xt
1.5 = 0 + (2.22 x 10³)(t)
t = 6.757 x 10^-4 s.

And plugged it into the distance traveled:
x = x₀ + v_0xt + 1/2 a_xt²
x = 0 + 0 + 1/2 (2.22 x 10³)(6.757 x 10^-4)²
x = 5.02 x 10^-4 m.

The book's answer said that it's 506 m so the ship will hit the reef, and the speed at which the ship hits the reef is 0.17 m/s, so the oil should be safe.
But I don't know how to get to the correct answers. :(

 

[phz]

The Attempt at a Solution

I don't exactly know what to do first, so I first found the acceleration of the ship's engines.
a = f/m = 8.0 x 10⁴N / 3.6 x 10⁷ kg = 2.22 x 10³ m/s²

Is it reasonable that the ship will receive an acceleration of 2220 m/s²? No, it's certainly not. Check your calculation in this step.

 

[shanu_bhaiya]

a = f/m = 8.0 x 10⁴N / 3.6 x 10⁷ kg = 2.22 x 10³ m/s²

shouldn't it be:
a = f/m = 8.0 x 10⁴N / 3.6 x 10⁷ kg = 2.22 x 10^-3 m/s²