two ends of a train moving with constant acceleration pass a certain point with velocities u and v. what is the velocity with which the middle point of the train passes the same point?
I have asked this question before
you have answered u+v/2. but my mam says it is (u^2+v^2/2)^1/2
please tell me how it comes.
I agree with mam. What you are figuring is the rms velocity.
Here is why:
velocity u is front, then velocity rear, u is
u^2=v^2+2aL where L is the length of the train. so
l= (u^2-v^2)/2a
now at midpoint,
m^2=v^2+2a L/2
m^2=v^2+a (u^2-v^2)/2a
=(u^2+v^2)/2
m= your answer.
Sorry for the earlier post.
we know that v2-u2=2as
NOTE :[2 behind u or v is power of them]
u= head of train
3u= tail of train [v=3u]
so (3u)2-u2=2as
9u2-u2=2as
8u2=2as
a=4u2/s
Now at middle of train s'=s/2
v'2-u2=2as'
v'2=u2+2as/2
put the value of a
v'2= u2+2*4u2/s*s/2
v'2=5u2
v'= root5u
siddharth sharma why are you taking taking tail of train as 3u
Let velocity at mod point is V .
<---------L------->.<---------L ---------->
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ {train}
v. V. u
(V^2-v^2)/2a=(u^2-V^2)/2a
So at midpoint velocity V = √{(u^2+v^2)/2}
To understand why your mam's answer is correct, let's break down the problem and work through it step by step.
Let's assume that the distance between the two ends of the train is L and the time taken to cover this distance is t.
First, let's calculate the average velocity of the train using the formula:
Average Velocity = (Total Displacement) / (Total Time)
The total displacement of the train is L, and the total time taken is t, so the average velocity is L/t.
Now, let's calculate the average acceleration of the train using the formula:
Average Acceleration = (Change in Velocity) / (Total Time)
The change in velocity is v - u (as one end of the train passes with velocity u and the other end passes with velocity v), and the total time taken is t. So the average acceleration is (v - u) / t.
Since the train is moving with constant acceleration, we can use the following equation of motion:
v = u + a*t
Where:
v is the final velocity (the velocity at which the second end of the train passes the point)
u is the initial velocity (the velocity at which the first end of the train passes the point)
a is the constant acceleration of the train
t is the total time taken
Rearranging the equation, we get:
v = u + a*t
Substituting the expression we found for average acceleration (a = (v - u) / t), we get:
v = u + ((v - u) / t) * t
Simplifying, we have:
v = u + v - u
v = v
This tells us that the final velocity of the second end of the train is the same as the initial velocity of the second end, which makes sense since the train is moving with constant acceleration.
Now, let's find the velocity with which the middle point of the train passes the same point. The middle point of the train is located at a distance of L/2 from either end.
We know that the average velocity of the train is L/t. And since the middle point of the train travels half the distance (L/2) in the same time (t), its average velocity is (L/2) / t.
To find the final velocity of the middle point, we can use the formula:
Final Velocity = (Initial Velocity) + (Acceleration) * (Time)
In this case, the initial velocity is u and the acceleration is (v - u) / t. The time taken for the middle point is the same as the time taken for the entire train, which is t.
Substituting the values in the formula, we get:
Final Velocity = u + ((v - u) / t) * t
Simplifying, we have:
Final Velocity = u + (v - u)
Final Velocity = v
So, the final velocity of the middle point of the train is v, which is the same as the final velocity of the second end of the train.
Therefore, the correct answer is v, not (u+v/2).
Let velocity at mod point is V .
<---------L------->.<---------L ---------->
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ {train}
v. V. u
(V^2-v^2)/2a=(u^2-V^2)/2a
So at midpoint velocity = √{(u^2+v^2)/2}