A train has a mass of 5.22 multiplied by 106 kg and is moving at 82.0 km/h. The engineer applies the brakes, resulting in a net backward force of 1.87 multiplied by 106 N on the train. The brakes are held on for 21.0 s.
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It is possible to use your data to compute a deceleration rate, and the velocity change after 21 s of braking.
a = F/m = 1.86*10^6/5.22*10^6
= 0.3563 m/s^2.
That is the deceleration rate.
After 21.0 s, the velocity will be lowered by a*t = 7.48 m/s
The original speed was 82.0 km/h = 22.78 m/s.
The final speed is 15.30 m/s.
That can be converted back to km/h if you wish.
Well, it sounds like the train is really applying the brakes! I hope it doesn't get a speeding ticket. Anyway, let me calculate the acceleration for you.
To find the acceleration, we can use Newton's second law of motion: force equals mass times acceleration (F = ma). In this case, the force is 1.87 * 10^6 N and the mass is 5.22 * 10^6 kg. We can rearrange the equation to solve for acceleration: a = F/m.
So, plugging in the numbers, we get:
a = (1.87 * 10^6 N) / (5.22 * 10^6 kg)
Calculating that, we find that the acceleration of the train is approximately 0.357 m/s^2.
Now, we can use the kinematic equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
The initial velocity of the train is 82.0 km/h, but we need to convert that to m/s first. There are 1000 meters in a kilometer and 3600 seconds in an hour, so 82.0 km/h is equal to (82.0 * 1000 m) / (3600 s). Plug that into the equation with the other known values:
v = (82.0 * 1000 m) / (3600 s) + (0.357 m/s^2)(21.0 s)
Calculating that, we find that the final velocity of the train is approximately 24.3 m/s.
So, after applying the brakes for 21.0 seconds, the train's velocity will drop to approximately 24.3 m/s. That's quite a slow down! Just like a clown tripped on a banana peel.
To calculate the change in velocity of the train, we can use the formula:
Change in velocity = (Net backward force * time) / mass
Given:
Net backward force (F) = 1.87 * 10^6 N
Time (t) = 21.0 s
Mass (m) = 5.22 * 10^6 kg
Plugging in the values into the formula, we get:
Change in velocity = (1.87 * 10^6 N * 21.0 s) / (5.22 * 10^6 kg)
Calculating the numerator:
(1.87 * 10^6 N * 21.0 s) = 39.27 * 10^6 Ns
Calculating the denominator:
(5.22 * 10^6 kg)
Now, let's divide the numerator by the denominator:
Change in velocity = 39.27 * 10^6 Ns / 5.22 * 10^6 kg
Simplifying:
Change in velocity = 7.52 m/s
Therefore, the change in velocity of the train is 7.52 m/s.
To find the change in velocity of the train after the brakes are applied, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration:
Net force = mass * acceleration
In this case, the net force on the train is given as 1.87 * 10^6 N and the mass of the train is 5.22 * 10^6 kg. We can rearrange the formula to solve for acceleration:
Acceleration = Net force / mass
Plugging in the values, we have:
Acceleration = (1.87 * 10^6 N) / (5.22 * 10^6 kg)
Acceleration = 0.358 m/s²
Now, we can use the kinematic equation to find the change in velocity of the train. The kinematic equation is:
Δv = a * t
where Δv is the change in velocity, a is acceleration, and t is the time. In this case, we're given that the brakes are held on for 21.0 s. Plugging in the values, we have:
Δv = (0.358 m/s²) * (21.0 s)
Δv = 7.50 m/s
Therefore, the change in velocity of the train after the brakes are applied is 7.50 m/s.