12.

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
options:
1.y=x
2.y=3x

X distance moved in first 2 secs.

=> x=0.2+(1/2).a.2^2= 2a ...(1)

velocity at the end of 2 secs.(say v)
=> v = 0+a.2 = 2a ....(2)

y distance covered in the next 2 secs.
=> y = v.t +(1/2).a.t^2
or y = 2a.2 + (1/2).a.2^2 = 6a ...(2)
From, (1) & (2) y = 3x ...Option (2)

a ball dropped on to the floor from a height of 10mrebounds to a height 2.5m. If the ball is in contact with the floor for 0.02s,its avg. acceleration during contact

To solve this problem, we need to use the equations of motion for uniformly accelerated motion.

The equations of motion are:
1. Distance (s) = ut + (1/2)at^2,
2. Final velocity (v) = u + at,
3. Final velocity squared (v^2) = u^2 + 2as,

where:
s = distance traveled,
u = initial velocity,
a = acceleration,
t = time taken.

Given that the particle starts from rest, the initial velocity (u) is zero.

Let's calculate the distance traveled in the first 2 seconds (x) using the equation of motion (1):
x = (1/2)at^2.

Given that time (t) is 2 seconds, we have:
x = (1/2) * a * 2^2
x = 2a.

Now, let's calculate the distance traveled in the next 2 seconds (y) using the equation of motion (1):
y = (1/2)at^2.

The total time taken for both distances is 4 seconds, so the time for the second interval is 2 seconds. Therefore:
y = (1/2) * a * 2^2
y = 2a.

Comparing the distances obtained:
x = 2a,
y = 2a.

We can see that y is equal to x. Therefore, the correct option is 1. y = x.