modern oil tankers weigh more than a half-million tons and have lengths of up to one-fourth mile. Such massive ships require a distance of 5.0 km (about 3.0 mi) and a time of 20 min to come to a stop from a top speed of 30 km/h.
What is the magnitude of a such a ship's average acceleration in m/s^2 in coming to a stop?
what is the magnitude of the ships average velocity in m/s?
Where did you get 3600 seconds?
why is it not 30-0/20=1.5
and
5/20=4 km/min = 4,000 m/min = 66.67 m/s?
5/20=4 km/min = 4,000 m/min = 66.67 m/s?
BECAUSE
5/20 = 1/4 not 4
30,000 m / 3600 s = 8.333 m/s
To calculate the magnitude of the ship's average acceleration in m/s^2, we first need to determine its initial velocity (vⱼ) and its final velocity (vᵢ) during the time it takes to come to a stop.
Given:
Initial velocity (vⱼ) = 30 km/h
Final velocity (vᵢ) = 0 km/h
We know that acceleration (a) is defined as the change in velocity over time. In this case, our initial velocity is greater than our final velocity, so the change in velocity is negative.
We also know that the time taken to stop (t) is 20 minutes, or 1200 seconds.
Let's first convert the velocities from km/h to m/s:
Initial velocity (vⱼ) = 30 km/h * (1000 m/1 km) * (1 h/3600 s) = 8.33 m/s
Final velocity (vᵢ) = 0 km/h * (1000 m/1 km) * (1 h/3600 s) = 0 m/s
Now, we can calculate the average acceleration (a):
a = (vᵢ - vⱼ) / t
= (0 m/s - 8.33 m/s) / 1200 s
= -8.33 m/s / 1200 s
≈ -0.0069 m/s²
Therefore, the magnitude of the ship's average acceleration in coming to a stop is approximately 0.0069 m/s².
Now let's calculate the magnitude of the ship's average velocity (V_avg) in m/s.
Average velocity is defined as the total displacement (Δx) divided by the time taken (t).
Given:
Distance (Δx) = 5.0 km = 5000 m
Time (t) = 20 min = 1200 s
Average velocity (V_avg) = Δx / t
= 5000 m / 1200 s
= 4.17 m/s
Therefore, the magnitude of the ship's average velocity in coming to a stop is approximately 4.17 m/s.
I figured it out. 3600 seconds in an hour.
5000 meters in 20*60 = 1200 seconds
Vi = 30,000 m / 3600 s = 83.3 m/s
v = Vi + a t
0 = 83.3 + a (1200)
a = - .0694 = - 6.94 * 10^-2 m/s^2
average velocity with constant acceleration = (first + last)/2
= 83.3/2 = 41.7 m/s