A train accelerates from the rest at the station A at a rate of 0.8 m/s2 for 25 s, the travels at constant velocity for 77.5 s before it comes to rest at the station B after period of retardation lasting 20 s. What is the distance between stations A and B, and the average velocity of the train between the stations?
I have calculated the distance of first part via equation s=vit+1/2at^2 and got the answer 250 then i found the velocity
20 m/s and now got stuck with retardation part . Please tell me ! soon
after 25 seconds, v=25*.8 = 20
when it starts to decelerate, a = -1 m/s^2, since it takes 20s to decrease v to 0.
so, the total distance is
s = 1/2 .8*20^2 + 20*77.5 + 20*10 - .5*400 = 1710 m
avg speed is thus 1710/(20+77.5+20) m/s
To calculate the distance traveled during the retardation period, we first need to find the final velocity of the train at the end of the constant velocity period.
Given:
Acceleration during acceleration period (a₁) = 0.8 m/s²
Time during acceleration period (t₁) = 25 s
To find the final velocity (v) at the end of the constant velocity period, we use the formula:
v = u + at
where u is the initial velocity (which is also the final velocity from the acceleration period, since the train continues at constant velocity).
Initial velocity at the end of the acceleration period (u) = 20 m/s (as you mentioned)
Using the formula, we can find the final velocity (v):
v = u + at
v = 20 + 0.8 × 25
v = 20 + 20
v = 40 m/s
Now, during the retardation period, the train is coming to a stop. Since the retardation is the negative acceleration, we can use the same formula as in the acceleration period:
s = ut + (1/2)at²
Given:
Retardation (a₂) = -0.8 m/s² (notice the negative sign)
Time during retardation period (t₂) = 20 s
Initial velocity at the start of the retardation period (u₂) = 40 m/s (obtained from the constant velocity period)
Using the formula, we can find the distance (s) during the retardation period:
s = u₂t₂ - (1/2)a₂t₂²
s = 40 × 20 - (1/2)(-0.8)(20)²
s = 800 - (1/2)(-0.8)(400)
s = 800 - (-0.4 × 400)
s = 800 - (-160)
s = 800 + 160
s = 960 m
Therefore, the distance between stations A and B is 250 m (from the acceleration period) + 960 m (from the retardation period), which equals 1210 m.
To find the average velocity (v_avg) between stations A and B, we use the formula:
v_avg = total distance / total time
Given:
Total distance = 1210 m (calculated above)
Time during the acceleration period (t₁) = 25 s
Time during the constant velocity period (t) = 77.5 s
Time during the retardation period (t₂) = 20 s
Total time (T) from A to B = t₁ + t + t₂
T = 25 + 77.5 + 20
T = 122.5 s
Using the formula, we can find the average velocity (v_avg):
v_avg = total distance / total time
v_avg = 1210 / 122.5
v_avg = 9.877 m/s (approximately)
Therefore, the average velocity of the train between stations A and B is approximately 9.877 m/s.
To calculate the distance of the retardation part, you can use the equation s = vit + 1/2at^2, where s is the distance, vi is the initial velocity, t is the time, and a is the acceleration.
Given that the train comes to rest at station B after a period of retardation lasting 20 s, we can assume the final velocity at station B is 0 m/s.
The initial velocity at station B (before the period of retardation) is the same as the final velocity at station A, which is 20 m/s (which you have correctly calculated).
Now, using the equation s = vit + 1/2at^2, we can plug in the known values:
- vi = 20 m/s (initial velocity at station B)
- t = 20 s (time of retardation)
- a = acceleration (which we need to find)
Since the train comes to rest, the final velocity is 0 m/s, so we can rearrange the equation to solve for acceleration (a):
0 = 20 + a * 20
Solving for a, we get:
-20 = 20a
Dividing both sides by 20:
-1 = a
Therefore, the acceleration during the retardation phase is -1 m/s^2 (negative since it's deceleration or retardation).
Now we can calculate the distance (s) during the retardation phase using the same equation:
s = vit + 1/2at^2
s = 20 * 20 + 0.5 * (-1) * 20^2
s = 400 - 0.5 * 400
s = 400 - 200
s = 200 m
So, the distance between stations A and B is the sum of the distances in the acceleration and retardation phases:
Distance = Distance in acceleration phase + Distance in retardation phase
= 250 m + 200 m
= 450 m
Finally, to calculate the average velocity of the train between the stations, you can use the formula:
Average velocity = Total distance / Total time taken
In this case, the total distance is 450 m (as calculated) and the total time taken is 25 s (acceleration phase) + 77.5 s (constant velocity phase) + 20 s (retardation phase) = 122.5 s.
Therefore, the average velocity of the train between the stations is:
Average velocity = 450 m / 122.5 s
≈ 3.67 m/s (rounded to two decimal places)
So, the distance between stations A and B is 450 meters, and the average velocity of the train between the stations is approximately 3.67 m/s.