A particle starts moving from rest with uniform acceleration.It travels a distance x in first 2 sec and a distance y in the next 2 sec.Then Answer) y=3x
correct.
Let the initial velocity be 0. Then in three seconds it becomes 3m/s and again in next three seconds it becomes 6m/s. We know that acceleration here is 3m/s square. Hence in the first case....... If we put the formula as 2as=v square --u square
We will get 2*3*x=9--0
By solving we would get x=3/2
Similarly for second case...
2as=v square --u square. Here since the initial velocity is 3m/s(as the final velocity in first case is initial here) acceleration is 3m/ square
And final velocity is 6m/s.
By using same formula
2as=v square -u square
2*3*Y=36-9
Y. =(27/6) 9/2
When we compare we get
3/2*3=9/2
Hence 3X=Y
To understand the relationship between the distances x and y, we need to use the equations of motion. The equations of motion for a particle undergoing uniform acceleration are:
1. Position at time t: s = ut + 0.5at^2
2. Final velocity at time t: v = u + at
3. Distance traveled during time t: s = 0.5(u + v)t
Given that the particle starts from rest, the initial velocity u = 0. From the information provided, we have:
For the first 2 seconds:
s1 = x
t1 = 2
Using equation 1, we can find the acceleration a:
x = 0.5 * a * (2^2)
x = 2a
For the next 2 seconds:
s2 = y
t2 = 2
Using equation 3, we can find the distance traveled during the next 2 seconds:
y = 0.5(0 + v2) * 2
y = v2
Now, let's find the final velocity at the end of the first 2 seconds, which will be the initial velocity for the next 2 seconds. Applying equation 2:
v2 = 0 + a * 2
v2 = 2a
We can substitute the value of v2 in the equation for y:
y = v2
y = 2a
But we know that x = 2a. Substituting this into the equation gives us:
y = 2x
Therefore, the relationship between y and x is y = 2x or y = 3x since 2 is a multiple of 3.
So, the answer is y = 3x.