Question

Let the initial velocity of the particle is 5m/s.it is accelerating at a acceleration of 1m/s^2.its final velocity is 10m/s.then the distance traveled by this particle will be given by:

Answer 1

Answer:

$\bigstar{\bold{Distance\:=\:37.5\:m}}$

Explanation:

$\Large{\underline{\underline{\it{Given:}}}}$

• Initial velocity (u) = 5 m/s
• Final velocity (v) = 10 m/s
• Acceleration(a) = 1 m/s²

$\Large{\underline{\underline{\it{To\:Find:}}}}$

• Distance travelled (s)

$\Large{\underline{\underline{\it{Solution:}}}}$

→ By the third equation of motion,

  v² - u² = 2as

→ Substituting the given datas, we get

  10² - 5² = 2 × 1 × s

  s = 75/2

$\boxed{\bold{Distance\:=\:37.5\:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

The three equations of motion are:

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² = 2as