~I got a and b right but c i can't figure out
Freight trains can produce only relatively small acceleration and decelerations.
(a) What is the final velocity of a freight train that accelerates at a rate of 0.0600 m/s2 for 8.00 min, starting with an initial velocity of 5.00 m/s?
33.8 m/s
(b) If the train can decelerate at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?
61.45 s
(c) How far will it travel in each case?
m (distance traveled during acceleration)
m (distance traveled during deceleration)
Vf^2=Vi^2+2ad
in each case, you know vf, vi, a (one them it is negative, solve for d)
or, you can figure average velocity (avg velocity= 1/2(Vf+vi), and distance= avgvelocty*timeinseconds
i don't understand what ad is
To find the distance traveled during acceleration and deceleration, we can use the kinematic equation:
Distance = (Initial velocity * time) + (0.5 * acceleration * time^2)
For the distance traveled during acceleration:
(a) Given:
Initial velocity (u) = 5.00 m/s
Acceleration (a) = 0.0600 m/s²
Time (t) = 8.00 min = 8.00 * 60 = 480 seconds
Using the formula:
Distance = (5.00 * 480) + (0.5 * 0.0600 * 480^2)
Distance = 2400 + 0.5 * 0.0600 * 230400
Distance = 2400 + 0.5 * 13824
Distance = 2400 + 6912
Distance = 9312 meters
So, the distance traveled during acceleration is 9312 meters.
For the distance traveled during deceleration:
(b) Given:
Initial velocity (u) = 33.8 m/s (final velocity from part a)
Acceleration (a) = -0.550 m/s² (negative because it is deceleration)
We want to find the time (t).
Using the rearranged formula:
Distance = (Initial velocity * time) + (0.5 * acceleration * time^2)
0 = (33.8 * t) + (0.5 * -0.550 * t^2)
Rearranging the equation and solving for t using the quadratic formula:
-0.275t^2 + 33.8t = 0
By factoring out t:
t * (-0.275t + 33.8) = 0
The solutions are t = 0 (which we discard as it does not make sense in this context) and -33.8 / -0.275 = 122.91 seconds.
So, the distance traveled during deceleration is:
Distance = (33.8 * 122.91) + (0.5 * -0.550 * 122.91^2)
Distance = 4140.998 + 0.5 * -0.550 * 15101.4081
Distance = 4140.998 + -4140.998
Distance = 0 meters
Therefore, the distance traveled during deceleration is 0 meters.
To summarize:
(a) The distance traveled during acceleration is 9312 meters.
(b) The distance traveled during deceleration is 0 meters.