A train is traveling up a 5.2° incline at a speed of 3.09 m/s when the last car breaks free and begins to coast without friction.
(a) How long does it take for the last car to come to rest momentarily?
(b) How far did the last car travel before momentarily coming to rest?
Initial velocity, u = 3.09 m/s
Final velocity, v = 0 m/s
acceleration due to gravity, g = 9.8 m/s/s
Acceleration, a = -g sin(5.2°)
Distance before coming to rest, S
Time to come to rest = (v-u)/a
Using
v²-u² = 2as
S can be found
S=(v²-u²)/2a
thanks much! ^^
You're welcome!
Why did the last car break free? Did it have a hot date and couldn't wait any longer? I guess it just wanted to coast and enjoy the view! Let's calculate its journey.
(a) To calculate the time it takes for the last car to come to rest momentarily, we need to find the deceleration first. The deceleration can be calculated using the component of gravity acting opposite to the motion of the car. So, the deceleration is given by:
a = g * sin(θ) = 9.8 m/s^2 * sin(5.2°)
Remember, sin(θ) takes degrees, not radians. Now we can use the equation of motion to find the time it takes to come to rest:
v = u + at
Since the final velocity is 0 m/s (rest), the initial velocity is 3.09 m/s (given), and the acceleration (deceleration) is -a (negative because it's opposing motion), we can plug in the values:
0 = 3.09 - a * t
Solving for t, we can find the time it takes for the last car to come to rest momentarily.
(b) To find the distance the last car traveled before momentarily coming to rest, we can use the equation of motion:
s = ut + (1/2)at^2
Since the final velocity is 0 m/s, the initial velocity is 3.09 m/s, the acceleration is -a, and we already know the time it takes for the last car to come to rest (from part a), we can substitute the values and solve for s.
So, let's calculate the value and have a laugh along the way!
To solve this problem, we need to analyze the motion of the last car after it breaks free from the train. We can use the principles of kinematics to find the time it takes for the car to come to rest momentarily and the distance it travels before that happens.
(a) To find the time it takes for the last car to come to rest momentarily, we can use the kinematic equation that relates final velocity (vf), initial velocity (vi), acceleration (a), and time (t):
vf = vi + at
Since the car is coasting without friction, the only force acting on it is the component of gravity along the incline. We can find this force by using the formula:
F = m * g * sin(theta)
where m is the mass of the car, g is the acceleration due to gravity, and theta is the angle of the incline.
The acceleration of the car can be calculated using Newton's second law:
F = m * a
Combining the two equations, we have:
m * a = m * g * sin(theta)
Canceling out the mass, we get:
a = g * sin(theta)
Now we can substitute the acceleration value into the kinematic equation:
0 = 3.09 m/s + a * t
0 = 3.09 m/s + (g * sin(theta)) * t
Solving for t, we find:
t = -3.09 m/s / (g * sin(theta))
Substituting the given values, where g = 9.8 m/s^2 and theta = 5.2°, we can calculate the time it takes for the last car to come to rest momentarily.
(b) To find the distance the car travels before momentarily coming to rest, we can use another kinematic equation that relates displacement (s), initial velocity (vi), acceleration (a), and time (t):
s = vi * t + (1/2) * a * t^2
Since the car is starting from rest, the initial velocity (vi) is 0 m/s. The acceleration (a) is still given by a = g * sin(theta). Plugging in these values, we can calculate the distance traveled by the last car before momentarily coming to rest.
It's important to note that in both parts (a) and (b), we assume that there is no frictional force acting on the car during its motion.