A train has a mass of 5.22 multiplied by 106 kg and is moving at 70.0 km/h. The engineer applies the brakes, which results in a net backward force of 1.87 multiplied by 106 N on the train. The brakes are held on for 37.0 s.
(a) What is the new speed of the train?
(b) How far does it travel during this period?
A train has a mass of 5.22*106kg? That is a really lightweight train. Think about that. My pickup truck weighs much more than that.
Vf=vi+a t
change vi 70km/h to m/s
a= force/mass= -1.87*106N/(5.22*106)
solve for vf
distance: avg velocity*time= (vi+Vf)*t/2
To find the new speed of the train and the distance it travels during this period, we can use the equations of motion.
(a) To find the new speed of the train, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 70.0 km/h = 19.4 m/s (converting km/h to m/s)
a = net backward force = -1.87 x 10^6 N / 5.22 x 10^6 kg (dividing force by mass)
t = 37.0 s
Substituting the given values into the equation, we get:
v = 19.4 m/s + (-1.87 x 10^6 N / 5.22 x 10^6 kg) * 37.0 s
Calculating this expression, we get:
v = 9.76 m/s
Therefore, the new speed of the train is 9.76 m/s.
(b) To find the distance the train travels during this period, we can use the equation:
s = ut + 0.5at^2
Where:
s = distance
u = initial velocity
a = acceleration
t = time
Given:
u = 19.4 m/s
a = -1.87 x 10^6 N / 5.22 x 10^6 kg
t = 37.0 s
Substituting the given values into the equation, we get:
s = 19.4 m/s * 37.0 s + 0.5 * (-1.87 x 10^6 N / 5.22 x 10^6 kg) * (37.0 s)^2
Calculating this expression, we get:
s ≈ 339.57 m
Therefore, the train travels approximately 339.57 meters during this period.
To answer these questions, we can use Newton's second law of motion and the equation for uniform acceleration to calculate the new speed and the distance traveled by the train.
(a) To find the new speed of the train, we can use the equation:
v = u + at
Where:
v = final velocity (new speed)
u = initial velocity (initial speed)
a = acceleration
t = time
In this case, the initial velocity (u) is given as 70.0 km/h, which we need to convert to m/s. To convert km/h to m/s, we can use the conversion factor: 1 km/h = 1000 m/3600 s.
So, the initial velocity (u) in m/s is:
u = (70.0 km/h) * (1000 m/3600 s) = 19.4 m/s
Now, let's find the acceleration (a) using Newton's second law of motion:
F = ma
Where:
F = net force applied on the train
m = mass of the train
a = acceleration
In this case, the net force (F) applied on the train is given as 1.87 multiplied by 106 N, and the mass of the train (m) is given as 5.22 multiplied by 106 kg.
So, the acceleration (a) is:
a = F / m = (1.87 multiplied by 106 N) / (5.22 multiplied by 106 kg) = 0.358 m/s^2
Now, we can substitute the values of u and a into the equation to find the new speed (v):
v = u + at = 19.4 m/s + (0.358 m/s^2) * (37.0 s) = ?
Calculating this gives us:
v ≈ 32.2 m/s
So, the new speed of the train is approximately 32.2 m/s.
(b) To find the distance traveled by the train during this period, we can use the equation for distance traveled with uniform acceleration:
s = ut + (1/2)at^2
Where:
s = distance traveled
u = initial velocity (initial speed)
a = acceleration
t = time
We already know the values of u, a, and t. Let's calculate the distance (s):
s = ut + (1/2)at^2 = (19.4 m/s) * (37.0 s) + (1/2) * (0.358 m/s^2) * (37.0 s)^2 = ?
Calculating this gives us:
s ≈ 651.5 m
So, the train travels approximately 651.5 meters during this period.