A train starts from rest from a station with acceleration 0.2m/s2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s2 .If total time spend is half an hour, then distance between two stations is_ ?
For the life of me, I don't understand when the retarding force begins.
To find the distance between the two stations, we can use the equation of motion:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Let's break down the problem into two parts:
1. Train accelerating from rest:
In this part, the train starts from rest, so the initial velocity (u) is 0 m/s. The acceleration (a) is given as 0.2 m/s^2. We need to find the time (t) it takes to reach maximum speed.
Using the equation v = u + at, we can rearrange it to find time:
t = (v - u) / a
Since the train starts from rest, its initial velocity (u) is 0 m/s. Let's assume the time taken to reach maximum speed is t1.
t1 = v / a
2. Train decelerating to come to rest:
In this part, the train is decelerating, so the final velocity (v) is 0 m/s. The acceleration (a) is given as -0.4 m/s^2 (negative because it's deceleration). We need to find the time (t) it takes to come to rest.
Using the same equation v = u + at, we can rearrange it to find time:
t = (v - u) / a
Since the final velocity (v) is 0 m/s, let's assume the time taken to come to rest is t2.
t2 = -u / a
Given that the total time spent is half an hour, or 0.5 hours, we can relate the times t1 and t2:
t1 + t2 = 0.5
Substituting the equations for t1 and t2:
v / a - u / a = 0.5
Now, let's substitute the values given in the problem: a = 0.2 m/s^2 and a = -0.4 m/s^2.
v / 0.2 - u / -0.4 = 0.5
2v + u = 1
Now, let's consider the motion from the first station to the second station. The distance covered when accelerating from rest can be calculated using the equation of motion:
s1 = u * t1 + (1/2) * a * t1^2
Substituting the values, we get:
s1 = 0 * t1 + (1/2) * 0.2 * t1^2
s1 = 0.1 * t1^2
The distance covered when decelerating to come to rest can be calculated using the equation of motion:
s2 = v * t2 + (1/2) * a * t2^2
Substituting the values, we get:
s2 = 0 * t2 + (1/2) * -0.4 * t2^2
s2 = -0.2 * t2^2
Since the train comes to rest at the end, the total distance covered is the sum of s1 and s2:
s = s1 + s2
s = 0.1 * t1^2 - 0.2 * t2^2
Now we have two equations:
2v + u = 1 (Equation 1)
s = 0.1 * t1^2 - 0.2 * t2^2 (Equation 2)
To solve these equations, we need to find the values of v, t1, and t2. Let's substitute the values we know:
v has been given as the maximum speed reached by the train, so let's denote it as vmax.
v = vmax
t1 = vmax / 0.2 (from earlier)
t2 = -u / -0.4 (from earlier)
Now we can solve for t1 and t2 using Equation 1:
2vmax + u = 1
Substitute u = 0 and rearrange the equation:
2vmax = 1
vmax = 1/2
Now, substitute the value of vmax in t1 and t2:
t1 = (1/2) / 0.2
t1 = 2.5 seconds
t2 = 0 / -(-0.4)
t2 = 0 seconds
Now we can substitute the values of t1 and t2 in Equation 2 to find the distance s:
s = 0.1 * (2.5)^2 - 0.2 * (0)^2
s = 0.0625 km
Therefore, the distance between the two stations is 0.0625 km, or 62.5 meters.