Question

A particle at rest starts moving with uniform acceleration covers a distance 114 m in 8 th second . What is the acceleration of the particle ? ​

Answer 1

Answer:

Explanation:

S=u +a/2(2n-1)

114=0+a/2(2×8-1)

114×2/15=a

a=15.2

Answer 2

◾Provided :-

→initial velocity = 0

→distance covered in 8th second = 144m

♦ So to find out the distance covered in 8th sec we will Subtract total distance covered in 7 sec from total distance covered in total distance covered in 8 sec During the motion .

◾ Let the

→acceleration of particle be "a"

→Distance covered in 7 sec = $S_7$

→Distance covered in 8 sec = $S_8$

◾Then from

⏩$S = ut + \frac{1}{2} a t^2$

➡️$S_7 = (0)(7)+ \frac{1}{2} a (7)^2$

$S_7 = 0 + \frac{1}{2} a 49$

$S_7 = \dfrac{49}{2}$ .....(i)

➡️$S_8 = (0)(8) + \frac{1}{2} a (8)^2$

$S_8 = 0 + \frac{1}{2} a 64$

$S_8 = \dfrac{64a}{2}$ ...(ii)

◾Now from (ii) - (i) we get distance travelled in 8th sec

$\dfrac{64a}{2} - \dfrac{49a}{2} = 114$

$\dfrac{64a-49a}{2} = 114$

$\dfrac{15a}{2} = 114$

$15a = 228$

$a = \dfrac{228}{15}$

$a = \dfrac{76}{5}$

$a = 15.2 \: ms^{-2}$