Question

Freight trains can produce only relatively small accelerations and decelerations. What is the final velocity, in meters per second, of a freight train that accelerates at a rate of 0.065 m/s^2 for 9.75 min, starting with an initial velocity of 3.4 m/s? If the train can slow down at a rate of 0.625 m/s^2, how long, in seconds, does it take to come to a stop from this velocity? How far, in meters, does the train travel during the process described in part (a)? How far, in meters, does the train travel during the process described in part (b)?

Answer 1

The final velocity of the freight train is approximately 21.2 m/s. The train takes approximately 33.92 seconds to come to a stop. The train travels approximately 15,087 meters during the acceleration and 3,655 meters during the deceleration.

Step-by-step explanation:

To find the final velocity of the freight train, we can use the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the given values, we have:

v = 3.4 m/s + (0.065 m/s2)(9.75 min)(60 s/min)

Solving this equation, we get the final velocity of the freight train to be approximately 21.2 m/s.

To find the time it takes for the train to stop, we can use the same equation:

v = u + at

But this time, the initial velocity is the final velocity from part (a), and the acceleration is the deceleration rate of 0.625 m/s2.

Plugging in the given values, we have:

0 = 21.2 m/s + (0.625 m/s2)t

Solving this equation, we find that it takes approximately 33.92 seconds for the train to come to a stop.

To find the distance traveled during the acceleration, we can use the equation:

s = ut + (1/2)at2

where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the given values, we have:

s = 3.4 m/s(9.75 min)(60 s/min) + (1/2)(0.065 m/s2)(9.75 min)(60 s/min)2

Solving this equation, we find that the train traveled approximately 15,087 meters during the acceleration.

To find the distance traveled during the deceleration, we can use the same equation:

s = ut + (1/2)at2

But this time, the initial velocity is the final velocity from part (a), and the acceleration is the deceleration rate of 0.625 m/s2.

Plugging in the given values, we have:

s = 21.2 m/s(33.92 s) + (1/2)(-0.625 m/s2)(33.92 s)2

Solving this equation, we find that the train traveled approximately 3,655 meters during the deceleration.

Answer 2

Step-by-step explanation:

Initial velocity of train u = 3.4 m⁻¹ , Acceleration a = .065 ms⁻² ,

time t = 9.75 x 60 = 585 s.

v = u + at

= 3.4 + .065 x 585

= 41.425 m / s

distance travelled during the acceleration ( s )

s = ut + 1/2 at²

= 3.4 x 585 + .5 x .065 x 585²

= 1989 + 11122.31

= 13111.31 m .

again for slowing process

u = 41.425 m / s

v = u -at

0 = 41.425 - 0.625 t

t = 66.28 s

v² = u² - 2as

0 = 41.425² - 2 x .625 s

s = 1372.82 m