Question

Moving with uniform acceleration from the state of rest a body covers 22m during the 6th sec of itz motion .what is distance covered by if during the first 6 sec?...plzzz with formula..explanation..

Answer 1

snth=u+a/2(2n-1)
Use this formula
22=0+a/2(2×6-1)
22=11a/2
44=11a
a=4 m/s^2
now putting this ,
s=ut+1/2at^2
s=(1/2)×4×36
s=2×36=72m ans

Answer 2

Answer:

The distance covered by if during the first 6 sec is 72 m

Explanation:

From the above question,

They have given :

Moving with a uniform acceleration from the state of rest a body covers 22m during the 6th sec of its motion.

Here we need to find the distance covered by if during the first 6 sec.

Let's first calculate the acceleration of the body.

We comprehend that the physique strikes with uniform acceleration. Therefore, the acceleration will stay the equal at some point of the motion.

Let a be the acceleration of the body, and t be the time taken to cowl a distance of 22m at some stage in the sixth 2d of its motion.

We can use the following kinematic equation to discover the acceleration:

          s = ut + (1/2)at^2

Here, u = zero (as the physique begins from rest), s = 22m, and t = 6s.

Therefore, 22 = zero + (1/2) * a * 6^2

Solving for a, we get a = 2.44 m/s^2

Now, the use of the method to locate distance included via the physique in the first 6 seconds of its motion:

          s = ut + (1/2)at^2

Here, u = 0, t = 6s, and a = 2.44 m/s^2

          t = 6

          s = $\frac{1}{2}$ a$t^{2}$

          s =  $\frac{1}{2}$ 4 x 6 x 6

          s = 72 m

The distance covered by if during the first 6 sec is 72 m

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