Science, asked by gourabbiswas81, 7 months ago

একটি কণা সরলরেখা বরাবর গতিশীল কণাটির যাত্রারম্ভের অবস্থান থেকে
3 s পরে অতিক্রান্ত দূরত্ব 180 cm এবং পরের 5 s সময়ে অতিক্রান্ত দূরত্ব
220 cm | কণাটির 9 s পর বেগ কত হবে?

Answers

Answered by ishetashukla2007

Explanation:

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Answered by knjroopa

Explanation:

Given From the starting position of a moving particle along a particle straight line .The distance covered after 3 s is 180 cm and the distance after 5 s 220 cm What will be the velocity of the particle after 9 s?

We have the equation of motion
S(t) = ut + ½ at^2
So u is initial velocity and a is acceleration
S(3) = u (3) + ½ x a x 3^2
180 = 3u + 9a / 2
6u + 9a / 2 = 180
6u + 9a = 360 ----------- 1
So we have
S(8) – S(3) = 220
8 x U + ½ x a x 8^2 = 220 + 180
8u + 32 a = 400 ------------ 2
So from 1 and 2 we have
6u + 9a = 360
8u + 32a = 400
So 2u + 3a = 120
2u + 8a = 100
Subtracting we get
So – 5a = 20
Or a = - 20 / 5
Or a = - 4 cm/s^2
2u + 3(-4) = 120
2u – 12 = 120
2u = 132
u = 132 / 2
u = 66 cm/s
Now velocity at 9 secs will be
V = u + at
= 66 + (-4) 9
= 66 – 36
V = 30 cm / s

Reference link will be

https://brainly.in/question/18574191

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