Question

solve and verify the answer 2y -2/9 =4y-4/3​

Answer 1

$\huge\bold\red{Given:-}$

Velocity of a car started from rest reaches 24 m/s in 4s

$\huge\bold\pink{To~find:-}$

$\sf{Initial ~velocity}$

$\sf{Final ~velocity}$

$\sf{Acceleration}$

$\huge\bold\purple{SolutiØn:-}$

$\sf{Initial~velocity=0~{started~from~rest}}$

$\sf{Final~velocity=24~m/s}$

$\sf{Time=4s}$

$\bold\orange{According~to~the~first~equation ~of ~the~motion:-}$

$\longrightarrow$$\sf\green{v=u+at}$

Where “v” is final velocity “u” is initial velocity, “a” is acceleration and “t” is time taken.

$\huge\bold\pink{Put~the~values:-}$

$\longrightarrow$ $\sf{24=0+a×4}$

$\longrightarrow$ $\sf{a=}$$\frac{24}{4}$

$\longrightarrow$ $\sf{a=6~m/s²}$

$\huge\mathfrak\purple{Hence,}$

Acceleration of the car is 6 m/s²

$\huge{\underline{\texttt{\pink{More~to~know→}}}}$

$\sf{s=ut+at²(second~equation~of~motion)}$

$\sf{v²=u²+2as(Third~equation~of~motion)}$

Negative value of acceleration is known as retardation.

Answer 2

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Question :

Solve and verify the answer :

$\\ \red \bigstar \rm \: 2y - \frac{2}{9} = 4y - \frac{4}{3}$

To find :

$\red \bigstar \rm \: the \: value \: of \: \: y \:$

Solution :

$\\ \implies \large \sf \: 2y - \frac{2}{9} = 4y - \frac{4}{3}$

$\\ \implies \large \sf \: 2y \times 9 - 2 = 4y \times 9 - \frac{4}{3} \times 9$

$\\ \implies \large \sf \: 18y - 2 = 36y - 12$

$\\ \implies \large \sf \: 18y - 36y - 2 + 12 = 0$

$\\ \implies \large \sf \: - 18y + 10 = 0$

$\\ \implies \large \sf \: - 18y = - 10$

$\\ \implies \large \boxed{ \sf{ \red \bigstar \: y = \frac{5}{9} }}$

The answer will be y=5/9