Question

A body initially at rest, moves with an acceleration of 5 m/s2 for 5 sec. Find the distance travelled in:- 4 seconds 5 seconds 5th second

Answer 1

Answer:

distance traveled in 5th second = Distance traveled in 5 s− Distance traveled in 4 s

Using s=ut+(1/2)at^2

 

Distance traveled in 5 s =(1/2)×5×5  ^2

 

                                         =62.5 m

Distance traveled in 4 s =(1/2)×5×4  ^2

 

                                         =40 m

Distance traveled in 5th s =(1/2)×5×5  ^2

 

                                            =62.5 m=S2  −S 1

​  

=(62.5−40)m=22.5m

                                 

Explanation:

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Answer 2

GiveN :-

• A particle initially at rest, moves with an acceleration 5m/s² for 5 s .

To FinD :-

• The distance travelled by the body in 4 seconds.
• The distance travelled by body in 5th s .

SolutioN :-

Given that , A particle initially at rest, moves with an acceleration 5m/s² for 5 s. And we need to find the distance travelled by the body in 4 seconds and 5th second So here ,

$\red{\frak{ Given }}\begin{cases} \textsf{ Initial velocity of particle =\textbf{ 0m/s } .}\\\textsf{ Acceleration of the particle = \textbf{ 5m/s}.}^2 \\\textsf{ Time of travelling =\textbf{ 4 seconds} .}\end{cases}$

★ Using the second equⁿ of motion :-

$\sf:\implies\pink{ s = ut +\dfrac{1}{2}at^2}\\\\\sf:\implies s = 0(4) + \dfrac{1}{2} \times 5m/s^2\times (4s)^2 \\\\\sf:\implies s = 0 + 5 \times \dfrac{16}{2} m \\\\\sf:\implies s = 5 \times 8 m \\\\\sf:\implies\underset{\blue{\sf Required \ Distance}}{\underbrace{\boxed{\pink{\frak{ Distance = 40 \ m }}}}}$

$\rule{200}2$

★ Using the formula of nth second :-

$\sf:\implies \pink{ S_n = u +\dfrac{1}{2}a [ 2n - 1 ] } \\\\\sf:\implies S_5 = 0 + \dfrac{1}{2}\times 5m/s^2 [ 2(5) - 1 ] \\\\\sf:\implies S_5 = 2.5 \times [ 10 - 1 ] m \\\\\sf:\implies S_5 = 2.5 \times 9 m \\\\\sf:\implies \underset{\blue{\sf Required \ Distance}}{\underbrace{\boxed{\pink{\frak{ Distance_5 = 22.5 \ m }}}}}$