A freight train has a mass of 1.5 x 10^7 kg . If the locomotive can exert a constant pull of 7.5 X 10^5 N, how long does it take to increase the speed of the train from rest to 80 km /h?

444s

To calculate the time it takes to increase the speed of the train from rest to 80 km/h, we need to use Newton's second law of motion, which states that the force applied to an object is equal to its mass multiplied by its acceleration:

F = m * a

In this case, the force (F) exerted by the locomotive is given as 7.5 x 10^5 N, and the mass (m) of the train is given as 1.5 x 10^7 kg. We need to find the acceleration (a) of the train.

Next, we can use the formula for calculating acceleration:

a = ∆v / ∆t

where ∆v is the change in velocity and ∆t is the change in time.

Given that the train starts from rest (0 km/h) and increases to 80 km/h, the change in velocity (∆v) is 80 km/h - 0 km/h = 80 km/h.

However, it's necessary to convert the velocity from km/h to m/s, as the SI unit of acceleration is meters per second squared (m/s²).

To convert km/h to m/s, we use the conversion factor: 1 km/h = 1000 m/3600 s = 5/18 m/s.

Therefore, the change in velocity (∆v) is (80 km/h) * (5/18 m/s) = (400/18) m/s ≈ 22.22 m/s.

Now, we can rearrange the equation for acceleration to solve for the change in time (∆t):

∆t = ∆v / a

Plugging in the values, we get:

∆t = 22.22 m/s / a

Now, we can solve for acceleration (a) using Newton's second law:

F = m * a
a = F / m

Plugging in the values, we get:

a = (7.5 x 10^5 N) / (1.5 x 10^7 kg)
a ≈ 0.05 m/s²

Now, we can substitute the value of acceleration (a) into the equation for ∆t:

∆t = 22.22 m/s / 0.05 m/s²
∆t ≈ 444.4 s

So, it takes approximately 444.4 seconds (s) for the train to increase its speed from rest to 80 km/h.

To determine the time it takes to increase the speed of the train from rest to 80 km/h, we can use Newton's second law of motion. The formula for Newton's second law is:

F = ma

Where:
F = force applied (in Newtons)
m = mass of the object (in kilograms)
a = acceleration (in meters per second squared)

In this case, the force is the pull exerted by the locomotive, which is 7.5 x 10^5 N, and the mass of the train is 1.5 x 10^7 kg. We can rearrange the formula to solve for acceleration:

a = F/m

Plugging in the values, we get:

a = (7.5 x 10^5 N) / (1.5 x 10^7 kg)

a = 0.05 m/s^2

Now, we need to convert the final speed from kilometers per hour to meters per second. There are 1000 meters in a kilometer and 3600 seconds in an hour, so:

Final speed = (80 km/h) * (1000 m/km) * (1 h/3600 s)

Final speed ≈ 22.22 m/s

To find the time it takes to achieve this final speed, we can use the following kinematic equation:

v = u + at

Where:
v = final velocity (22.22 m/s)
u = initial velocity (0 m/s)
a = acceleration (0.05 m/s^2)
t = time (in seconds)

Rearranging the equation to solve for time:

t = (v - u) / a

Plugging in the values, we get:

t = (22.22 m/s - 0 m/s) / 0.05 m/s^2

t ≈ 444.4 s

Therefore, it takes approximately 444.4 seconds to increase the speed of the train from rest to 80 km/h.