Question

A train is traveling up a 2.88 degree incline at a speed of 4.31 m/s when the last car breaks free and begins to coast without friction. (a) How much time 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?

Answer 1

The last car will not come to a complete rest momentarily on this incline. The distance traveled by the last car before momentarily coming to rest is approximately 3.84 meters.

Step-by-step explanation:

(a) To find the time it takes for the last car to come to rest momentarily, we can use the equation:
vf = vi + at

Where:
vf = final velocity = 0 m/s (since the car comes to rest momentarily)
vi = initial velocity = 4.31 m/s
a = acceleration = g * sin(θ) (where g is the acceleration due to gravity and θ is the incline angle)
t = time

Plugging in the values:

`0 = 4.31 + (9.8 * sin(2.88)) * t`

`0 = 4.31 + (9.8 * 0.0499) * t`
(9.8 * 0.0499) * t = -4.31
t = -4.31 / (9.8 * 0.0499)
t ≈ -0.89 seconds

Since time cannot be negative, the last car will never come to a complete rest momentarily on this incline.

(b) To find the distance traveled by the last car before momentarily coming to rest, we can use the equation:

`x = vit + 0.5at^2`

Plugging in the values:

`x = 4.31 * (-0.89) + 0.5 * (9.8 * sin(2.88))(-0.89)^2`

x = -3.84 meters

Therefore, the distance traveled by the last car before momentarily coming to rest is approximately 3.84 meters.

Answer 2

The last car's time to come to rest and the distance traveled before coming to rest on a 2.88-degree incline can be calculated using the equations of motion, considering the component of gravitational acceleration along the incline as the deceleration force.

Step-by-step explanation:

The scenario describes a physics problem involving a train car that breaks free and starts coasting up an incline without friction. To find the time it takes for the car to come to rest (momentarily stop), we need to apply the equations of motion under constant acceleration (in this case, deceleration due to gravity’s component along the incline). The deceleration 'a' is g*sin(θ) where g is the acceleration due to gravity (9.8 m/s2) and θ is the incline angle.

(a) To find the time to come to rest, use the equation v = u + at, where 'v' is the final velocity (0 m/s since it comes to rest), 'u' is the initial velocity (4.31 m/s), 'a' is the deceleration (-g*sin(θ)), and 't' is the time. Rearrange to solve for 't'.

(b) To find out how far the car coasted, use the equation s = ut + (1/2)at2, where 's' is the distance traveled. Once 't' is found, we can calculate 's' using the initial velocity and deceleration.