The engine of a locomotive exerts a constant force of 6.5 105 N to accelerate a train to 90 km/h. Determine the time taken for the train of mass 1.8 107 kg to reach this speed from rest.
Well, well, well, looks like the train wants to speed things up! Let's calculate the time it takes for this locomotive to turn from a snail into a cheetah!
To find the time taken for the train to reach a speed of 90 km/h, we can use Newton's second law, which states that force equals mass times acceleration. In this case, force is given as 6.5 x 10^5 N, and the mass of the train is 1.8 x 10^7 kg.
First, we need to convert the speed from km/h to m/s, because the SI unit system prefers the latter. So, 90 km/h is equal to 25 m/s.
Now, let's use the formula:
Force = mass x acceleration.
We rearrange the formula to solve for acceleration:
Acceleration = Force / mass.
Plugging in the values, we get:
Acceleration = 6.5 x 10^5 N / 1.8 x 10^7 kg.
Calculating that gives us:
Acceleration ≈ 3.61 m/s^2.
Now, to find the time, we can use the equation:
Time = (Final Velocity - Initial Velocity) / Acceleration.
The initial velocity here is 0 m/s because the train starts from rest, and the final velocity is 25 m/s.
Plugging in the values, we have:
Time = (25 m/s - 0 m/s) / 3.61 m/s^2.
Calculating that gives us:
Time ≈ 6.93 seconds.
Voila! It took approximately 6.93 seconds for the train to reach a speed of 90 km/h. Now, I hope it enjoys its new speedy journey!
To determine the time taken for the train to reach this speed, we can use Newton's second law of motion.
Step 1: Convert the speed from km/h to m/s.
1 km/h is equal to 1000 m/3600 s.
Therefore, 90 km/h = (90 * 1000) m/3600 s = 25 m/s.
Step 2: Use Newton's second law of motion to calculate the acceleration.
Newton's second law states that force (F) equals mass (m) multiplied by acceleration (a): F = m * a
Rearranging the formula, acceleration (a) equals force (F) divided by mass (m): a = F / m
Given force (F) = 6.5 * 10^5 N and mass (m) = 1.8 * 10^7 kg, we can substitute these values to find the acceleration (a):
a = (6.5 * 10^5 N) / (1.8 * 10^7 kg) = 3.61 m/s^2 (approximately).
Step 3: Use the kinematic equation to calculate the time taken.
The kinematic equation that relates initial velocity (v0), final velocity (v), acceleration (a), and time (t) is:
v = v0 + (a * t)
In this case, the initial velocity (v0) is 0 m/s (since the train starts from rest), the final velocity (v) is 25 m/s, and the acceleration (a) is 3.61 m/s^2.
Substituting these values into the kinematic equation, we have:
25 m/s = 0 m/s + (3.61 m/s^2 * t)
Rearranging and solving for time (t):
t = (25 m/s) / (3.61 m/s^2)
t ≈ 6.93 s
Therefore, the time taken for the train to reach a speed of 90 km/h from rest is approximately 6.93 seconds.
To determine the time taken for the train to reach the speed of 90 km/h from rest, we need to use Newton's second law of motion and the equations of motion.
First, let's convert the speed from km/h to m/s since the units need to be consistent for calculations. We know that 1 km/h is equal to 1000 m/3600 s.
So, the speed of the train, v = 90 km/h = (90 * 1000 m)/(3600 s) = 25 m/s.
Next, we'll use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the force exerted by the locomotive is given as 6.5 * 10^5 N.
Force (F) = mass (m) * acceleration (a)
Rearranging the equation, we get:
acceleration (a) = Force (F) / mass (m)
Substituting the given values:
a = 6.5 * 10^5 N / 1.8 * 10^7 kg = 0.036 m/s^2
Now, we can use one of the equations of motion that relates acceleration, initial velocity, final velocity, and time:
v = u + at
where:
v = final velocity = 25 m/s
u = initial velocity = 0 m/s (since the train is at rest)
a = acceleration = 0.036 m/s^2 (as calculated earlier)
t = time taken
Substituting the known values into the equation, we have:
25 m/s = 0 + 0.036 m/s^2 * t
Simplifying the equation:
25 m/s = 0.036 m/s^2 * t
Dividing both sides of the equation by 0.036 m/s^2:
t = 25 m/s / 0.036 m/s^2 = 694.44 s
Therefore, the time taken for the train to reach a speed of 90 km/h from rest is approximately 694.44 seconds.
6.5 * 10^5 I expect you mean and 1.8*10^7
F = m a
6.5 * 10^5 = 1.8*10^7 a
a = 3.61*10^-2 m/s^2
90 km/h (1 h/3600s)(1000 m/km)
= 25 m/s
v = a t
25 = 3.61*10^-2 t
t = 6.93 *10^2 s = 693 s = 11.5 min