Question

A pebble is released from a cliff and hits the earth surface with a speed of 50 m/s in
2.5 s. Find the acceleration​

Answer 1

AnswEr :

• Initial velocity (u) = 0 m/s
• Final velocity (v) = 50 m/s
• Time (t) = 2.5s

Acceleration : Rate of change of velocity.

$\\ \longrightarrow \tt{a\ =\ \dfrac{v\ -\ u}{t}} \\ \\ \\ \longrightarrow \tt{a\ =\ \dfrac{50\ -\ 0}{2.5}} \\ \\ \\ \longrightarrow \tt{a\ =\ \dfrac{50}{2.5}} \\ \\ \\ \longrightarrow \tt{a\ =\ \dfrac{500}{25}} \\ \\ \\ \large \longrightarrow \underline{\boxed{\frak{a\ =\ 20\ ms^{-2}}}}$

Answer 2

Given :

• Initial velocity (u) = 0 m/s
• Final velocity (v) = 50 m/s
• Time (t) = 2.5 seconds

To find :

• Acceleration (a)

According to the question,

→ v = u + at

Where,

• v stands for Final velocity
• u stand for Initial velocity
• t stands for Time
• a stands for Acceleration

Now,

Substituting the values,

→ 50 = 0 + a(2.5)

→ 50 - 0 = 2.5a

→ 50 = 2.5a

→ 50 ÷ 2.5 = a

→ 20 = a

∴ The acceleration is 20 m/s²...

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OTHER FORMULA :

Newton's first equation of motion :

• v = u + at

Newton's second equation of motion :

• v² = u² + 2as

Newton's third equation of motion :

• s = ut + ½ at²