Question

initial velocity = 0m/s distance =400m time=16sec find the acceleration
​

Answer 1

❒ Given :-

• ➊ Initial velocity of body = 0 m/s
• ➋ Distance travelled = 400 m
• ➌ Time taken = 16 sec

❒ To Find :-

• ➊ Acceleration of the body

❒ Solution :-

By using Law of motion

$\large \longmapsto \underline { \boxed{\red{ \bf s = ut + \frac{1}{2 } a {t}^{2}}}}$

Here :-

• s = Distance
• u = Initial velocity
• t = time
• a = acceleration

★ Substitute values in formula

$:\implies \sf400 = 0 \times 16 + \frac{1}{2} \times a \times {(16)}^{2}$

$:\implies\sf400 = \frac{1}{2} \times a \times 256$

$:\implies\sf400 = a \times 128$

$:\implies\sf a = \frac{400}{128}$

$:\implies\sf a = 3.125 \: {ms}^{ - 2}$

$\large\therefore \underline{\bf \green{Acceleration\: of \: body = 3.125 \: m/s}}$

Answer 2

Answer:

❒ Given :-

➊ Initial velocity of body = 0 m/s

➋ Distance travelled = 400 m

➌ Time taken = 16 sec

❒ To Find :-

➊ Acceleration of the body

❒ Solution :-

By using Law of motion

\begin{gathered} \large \longmapsto \underline { \boxed{\red{ \bf s = ut + \frac{1}{2 } a {t}^{2}}}} \\ \end{gathered}

⟼

s=ut+

2

1

at

2

Here :-

s = Distance

u = Initial velocity

t = time

a = acceleration

★ Substitute values in formula

\begin{gathered} :\implies \sf400 = 0 \times 16 + \frac{1}{2} \times a \times {(16)}^{2} \\ \end{gathered}

:⟹400=0×16+

2

1

×a×(16)

2

\begin{gathered}:\implies\sf400 = \frac{1}{2} \times a \times 256 \\ \end{gathered}

:⟹400=

2

1

×a×256

\begin{gathered}:\implies\sf400 = a \times 128 \\ \end{gathered}

:⟹400=a×128

\begin{gathered}:\implies\sf a = \frac{400}{128} \\ \end{gathered}

:⟹a=

128

400

\begin{gathered}:\implies\sf a = 3.125 \: {ms}^{ - 2} \\ \end{gathered}

:⟹a=3.125ms

−2

\begin{gathered}\large\therefore \underline{\bf \green{Acceleration\: of \: body = 3.125 \: m/s}} \\\end{gathered}

∴

Accelerationofbody=3.125m/s

Step-by-step explanation:

Hope it helps you