The Speed Of A Train Is Reduced From 80km/h to 40km/h In A Distance Of 500m On Applying The Brakes Calculate (I) How Much Further Will The Train Travel Before Coming To Rest (II) Assuming The Retardation Remain Constant,how Long Will It Take To Bring The Train To Rest After The Application Of The Brakes
V = 40km/h = 40,000m/3600s = 11.1 m/s.
Vo = 80km/h = 80,000m/3600s = 22.2 m/s.
1. V^2 = Vo^2 + 2a*d = (11.1)^2.
(22.2)^2 + 2a*500 = (11.1)^2,
1000a = (11.1)^2-(22.2)^2 = -369.63, a = -0.370 m/s^2.
(22.2)^2 + 2*(-0.370)*d = 0.
-0.740d = -(22.2)^2, d = 666 m. further.
2. V = Vo + a+t = 0.
22.2 - 0.370*t = 0, t = 60 s.
To solve this problem, we need to use the equations of motion. Let's denote the initial velocity of the train (u) as 80 km/h, the final velocity (v) as 40 km/h, the distance traveled (s) as 500 m, and the acceleration or retardation (a) as unknown.
Since the speed is reduced, we can use the equation:
(v^2 - u^2) = 2as
To calculate how much further the train will travel before coming to rest (I), we need to find the distance (s) when the final velocity (v) is zero. The equation becomes:
(0^2 - 80^2) = 2a(s)
-6400 = 2a(s)
Dividing both sides by 2a:
s = -6400 / (2a)
To calculate the time it takes to bring the train to rest (II), we need to find the time (t) when the final velocity (v) is zero. We can use the equation:
v = u + at
Since the final velocity v is zero:
0 = 80 + at
Rearranging the equation to solve for time t:
t = -80 / a
Therefore, to calculate the answer for (I) and (II), we need to find the value of acceleration (a).
Given that the distance traveled (s) is 500 m, we can substitute this value into equation (I):
500 = -6400 / (2a)
Multiplying both sides by 2a:
1000a = -6400
Dividing both sides by 1000:
a = -6.4 m/s^2
Now we can substitute this value for acceleration (a) into the equations for (I) and (II).
(I) How much further will the train travel before coming to rest?
s = -6400 / (2 * -6.4)
s ≈ 500 m
Therefore, the train will travel an additional 500 meters before coming to rest.
(II) Assuming the retardation remains constant, how long will it take to bring the train to rest after the application of the brakes?
t = -80 / -6.4
t ≈ 12.5 seconds
Therefore, it will take approximately 12.5 seconds to bring the train to rest after the application of the brakes.
To calculate how much further the train will travel before coming to rest, we need to find the total distance covered by the train during the deceleration. The initial speed of the train is 80 km/h, and the final speed is 40 km/h. We know the distance covered is 500 m.
To calculate the distance traveled during deceleration, we can use the equation:
v^2 = u^2 + 2as,
where v = final velocity, u = initial velocity, a = acceleration/retardation, and s = distance traveled.
Here, the initial velocity (u) of the train is 80 km/h (which needs to be converted to m/s). The final velocity (v) is 40 km/h (also needs to be converted to m/s). The distance covered (s) is 500 m.
Converting the velocities to m/s:
Initial velocity (u) = 80 km/h = (80 * 1000) / 3600 = 22.22 m/s
Final velocity (v) = 40 km/h = (40 * 1000) / 3600 = 11.11 m/s
Using the equation, we can rearrange it to solve for the distance (s):
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (11.11^2 - 22.22^2) / (2a)
Since we do not have the value of acceleration (a), we cannot directly calculate the distance. However, assuming the retardation remains constant, we can calculate it in the next step.
To calculate the time it will take to bring the train to rest after the application of brakes, we use the formula:
v = u + at,
where v = final velocity (0 m/s), u = initial velocity (22.22 m/s), a = acceleration/retardation, and t = time taken.
Rearranging the equation, we get:
t = (v - u) / a
Substituting the values:
t = (0 - 22.22) / a
Again, we need the value of acceleration (a) to calculate the time taken.
To get the values for the distance traveled and the time taken, we need the value of the retardation or acceleration. Please provide the value of acceleration or any additional information to solve the problem completely.