A body moves with uniform acceleration with

a distance x for 1st sec. What is relation b/w x & y?
A) y=24x
B) y=16x
C) y=12x
D) y=8x

What is y? x is distance. Where is acceleration here?

Y=12x

Hira, how is explanation for your answer?

To determine the relationship between the distance x for the first second and the value of y, we need to understand the concept of uniform acceleration.

Uniform acceleration means that the object's velocity is changing by the same amount in every second. In this case, let's assume the object's initial velocity is u and the acceleration is a.

Using the first equation of motion, we have:

v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time taken

Since the object moves with uniform acceleration, after 1 second, the final velocity (v) will be the initial velocity (u) plus the acceleration (a):

v = u + a(1) = u + a

Now, we need to find the relation between the distance x and the value of y. To do this, we can make use of the second equation of motion:

x = ut + (1/2)at^2

Rearranging the equation, we have:

x = ut + (1/2)at^2
x = u(1) + (1/2)a(1)^2
x = u + (1/2)a(1)^2
x = u + (1/2)a

Since we know that v = u + a, we can rewrite the equation as:

x = v - (1/2)a

Now, let's substitute v in terms of u and a:

x = (u + a) - (1/2)a
x = u + (1/2)a

So, we have found the relation between the distance x for the first second and the value of y, which is given by:

y = x

Therefore, the correct answer is:

D) y = 8x