Question

An oil-tanker at sea will often cut its engines many hours before it reaches port to allow the water to slow the ship down. How far (in km) does the tanker travel if it takes 5.47 hours to come to rest from 30.7 m/s? (Hint: what is the acceleration?)

Answer 1

307.29 in km does the tanker travel if it takes 5.47 hours to come to rest from 30.7 m/s.

First convert the speed to km/h: 30.7 m/s * (1 km / 1000 m) * (3600 s / 1 h) = 110.52 km/h

Since the deceleration is constant, we can use the one-variable equation: x = v_i * t + 1/2 * a * t^2

where x is the final distance, v_i is the initial speed, t is the time taken, and a is the acceleration.

We know v_i = 110.52 km/h, t = 5.47 h, and x = 0.

We need to find a.

Solving the equation above for a, we get: a = -2 * v_i / t = -2 * 110.52 km/h / 5.47 h = -39.82 km/h^2

Now, plug in all the values into the equation for x:

x = 110.52 km/h * 5.47 h + 1/2 * -39.82 km/h^2 * (5.47 h)^2

x ≈ 307.29 km

So the answer is 307.29