The Speed Of A Train Is Reduced From 80km/h To 40km/h In A Distance Of 500m On Applying The Brakes (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
I need the solution
To answer these questions, we need to use the kinematic equations of motion. The relevant equation in this case is:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, as the train comes to rest)
u = initial velocity (80 km/h or 22.22 m/s in this case)
s = distance (the unknown in the first question)
a = acceleration (retardation in this case)
Let's solve each question separately.
(I) How much further will the train travel before coming to rest?
We know that the initial velocity (u) is 22.22 m/s and the final velocity (v) is 0 m/s. We also have the acceleration (a), which we need to determine.
First, let's convert the final velocity from km/h to m/s:
0 km/h = 0 m/s
Now, let's rearrange the kinematic equation to solve for distance (s):
v^2 = u^2 + 2as
0^2 = (22.22)^2 + 2a(s)
0 = 493.0484 + 2as
2as = -493.0484
s = -493.0484 / (2a)
Since the retardation is negative (due to the braking action), we can rewrite the equation as:
s = 493.0484 / (2 |a|)
Given that s = 500 m, we can substitute this value into the equation:
500 = 493.0484 / (2 |a|)
Now, solve for |a|:
|a| = 493.0484 / (2*500)
|a| = 493.0484 / 1000
|a| = 0.4930484 m/s^2
Finally, substitute |a| back into the original equation to find s:
s = 493.0484 / (2 * 0.4930484)
s = 500 meters
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?
We already have the initial velocity (u = 22.22 m/s) and acceleration (retardation) |a| = 0.4930484 m/s^2. We need to find the time (t) it takes for the train to come to rest.
The equation to use is:
v = u + at
0 = 22.22 + (-0.4930484)t
Solve for t:
-0.4930484t = -22.22
t = -22.22 / (-0.4930484)
t = 45 seconds
Therefore, it will take 45 seconds for the train to come to rest after the application of the brakes, assuming the retardation remains constant.