A heavy freight train has a mass of 11800 metric tons. If the locomotive can pull with a force of 691000 N, how long does it take to increase it's speed from 0 to 70.3 km/h?
20.36=V0+5.9x10^-3
=349.1 s
is this correct?
No, not correct.
Your work is nonsense.
F=ma find a
knowing a, a=(Vf-Vo)/time
solve for time, but change Vf to m/s first.
oook but would that not be the same answer?
a=(20.36-0)/ t
t=20.36/.059
probably not how you rearrange it but all the other ways lead to numbers that are much too small
To determine the correct answer, we need to calculate the time it takes for the freight train to increase its speed from 0 to 70.3 km/h, given the mass of the train and the force applied by the locomotive.
First, let's convert the mass of the train from metric tons to kilograms:
11800 metric tons = 11800 * 1000 kg = 11,800,000 kg
Next, let's convert the speed from km/h to m/s:
70.3 km/h = 70.3 * (1000/3600) m/s = 19.5 m/s
Now, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = m * a), to calculate the acceleration of the train:
691,000 N = 11,800,000 kg * a
Solving for acceleration (a):
a = 691,000 N / 11,800,000 kg
a ≈ 0.0585 m/s^2
We can then use the following kinematic equation to find the time it takes for the train to reach a speed of 19.5 m/s:
v = v0 + a * t
Since the initial speed (v0) is 0 m/s, the equation becomes:
19.5 = 0 + 0.0585 * t
Solving for time (t):
t = 19.5 / 0.0585
t ≈ 333.33 seconds
Therefore, it takes approximately 333.33 seconds (or 333 seconds, rounding to the nearest whole number) for the train to increase its speed from 0 to 70.3 km/h.
So, the answer 349.1 seconds provided is not correct. The correct answer is approximately 333 seconds.