Question

a body poses in initial velocity of 5 metre per second and moves with uniform acceleration of 2 metre per calculate the distance moved by the body in 4 second​

Answer 1

Given

Initial velocity of body is 5 m/s, acceleration of body is 2 m/s². Time taken is 4 seconds.

To find

Distance covered by body.

Solution :

Here, initial velocity (u) = 5 m/s

Acceleration (a) = 2 m/s²

Time (t) = 4 seconds.

☯ USING 2ND EQUATION OF MOTION :

$\longrightarrow\underline{\boxed{\sf\blue{ s = ut + \dfrac{1}{2} at^2 }}}\: \star \\\\\longrightarrow\sf s = { (5 \times 4) + \dfrac{1}{\cancel{2}} \times \cancel{2} \times 4^2 } \\\\\longrightarrow\sf s = 20 + 16 \\\\\longrightarrow\underline{\boxed{\textsf{\textbf{s = 36 m }}}}\\\\\\\therefore\underline{\textsf{Distance covered by body = {\textbf{36 m }}}}$

$\rule{200}{2}$

$\qquad\underline{\underline{\text{Some more equations }}} : \\\\\dashrightarrow\rm{v = u + at \quad\big[1st \: equation\: of\: motion \big]} \\\\\dashrightarrow\rm{a = \dfrac{v - u}{t}} \\\\\dashrightarrow\rm{v^2 = u^2 + 2as \quad \big[3rd \: equation \: of \: motion \big]}$

Answer 2

Answer:

Distance is 36 m

Explanation:

Initial Velocity = 5m/s, Acceleration = 2m/s^2 and time = 4 sec

s = ut + 1/2 at^2 (Seconds Equation Of Motion)

s = 5(4) + 1/2 (2) (4)^2

s = 20 + 2/2 × 16

s = 20 + 1 × 16

s = 20 + 16

s = 36 m