Question

a vehicle moving with a velocity 10m/s attains a velocity 13m/s , after travelling a distance of 69 m. find its acceleration.​

Answer 1

Given,

u = 10m/s

v= 13m/s

s = 69m

We apply third eqn of motion :-

v² - u² = 2as

Putting values,

$(13) {}^{2} - (10) {}^{2} = 2 \times a\times 69$

$169 - 100 = 138 \times a$

a = 1/2

a = 0.5 m/s²

Answer 2

$\:\:\:\: \large \underline {\sf{Given\::}}$

• Initial velocity ( u ) = 10 m/s

• Final velocity ( v ) = 13 m/s

• Distance covered ( s ) = 69 m

$\:\:\:\: \large \underline {\sf{To\:Find\::}}$

• Acceleration ( a )

$\:\:\:\: \large \underline {\sf{Solution\::}}$

$\underline{\boxed{\sf{Formula\:used\:=\:{v}^{2}\:-\:{u}^{2}\:=\:2as}}}$

Now , substituting the values :

=> $\sf {13}^{2}\:-\:{10}^{2}\:=\:2(a)(69)$

=> $\sf 169\:-\:100\:=\:138(a)$

=> $\sf 69\:=\: 138(a)$

=> $\sf a\:=\: \dfrac{69}{138}$

=> $\sf a\:=\: \dfrac{1}{2}$

$\:\:\:\: \large \underline {\sf{\red{Acceleration\:=\:0.5\:m/s}}}$