Question

b) A body moving at 5cm/s is being accelerated uniformly
and passes over next 300 cm in 20 S. Calculate. 2
i) the acceleration of the body (ii) the velocity of the body
at the end of 20 S.
5cm/s বেগে চলমান একটি বস্তু সমত্বরণে ত্বরান্বিত হয়ে পরবর্তী 300
cm দূরত্ব 20 S এ অতিক্রম করল।
(ক) বস্তুটির ত্বরণ, (খ) 20 S পরে বস্তুটির বেগ নির্ণয় করাে।
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Answer 1

u = 5 cm/s

t = 20s

h = 300 cm

i)

H = ut + 1/2 at²

300 = 100 + 400a/2

200 = 200a

a = 1

acceleration = 1 cm/s²

ii)

u = 5 cm/s

a = 1 cm/s²

t = 20s

v = u + at

v = 5 + 20 × 1

v = 25

Velocity = 25 cm/s

Answer 2

$\huge {\blue{\underline{\underline{\mathfrak{AnSwEr :}}}}}$

$\small {\pink {\underline{\sf{Given :}}}}$

• Initial Velocity (u) = 5 m/s
• Distance or Height (s) = 300 cm
• Time (t) = 20 s

$\rule{200}{1}$

$\small {\orange{\underline{\sf{Solution :}}}}$

We have formula :

$\large \star{\boxed{\sf{s \: = \: ut \: \: + \: \dfrac{1}{2} at^2}}} \\ \\ \\ : \implies {\sf{300 \: = \: 5 \: \times \: 20 \: + \: \dfrac{1}{2} a(20)^2}} \\ \\ \\ : \implies {\sf{300 \: - \: 100 \: = \: \dfrac{1}{2}400a}} \\ \\ \\ : \implies {\sf{200 \: = \: 200a}} \\ \\ \\ : \implies {\sf{a \: = \: \dfrac{200}{200}}} \\ \\ \\ \Large {\boxed{\pink {\sf{Acceleration \: = \: 1 ms^{-2}}}}}$

$\rule{170}{0.5}$

Now, we have to find Final Velocity

We have formula :

$\large \star {\boxed{\sf{v \: = \: u \: + \: at}}} \\ \\ \\ : \implies {\sf{v \: = \: 5 \: + \: 1 \: \times \: 20}} \\ \\ \\ : \implies {\sf{v \: = \: 5 \: + \: 20}} \\ \\ \\ : \implies {\sf{v \: = \: 25 }} \\ \\ \\ \Large {\boxed{\red{\sf{Final \: Velocity \: = \: 25 \: ms^{-1}}}}}$