Question

a cyclist is travelling at 15 metre per second. she applies break so that she does not collide with a wall 18 metre away. what declaration must she have?

Answer 1

Answer:

$\bigstar{\bold{Deceleration=-6.25\:m/s^{2}}}$

Explanation:

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

• Initial velocity of cyclist (u) = 15 m/s
• Final velocity of cyclist (v) = 0 m/s
• Distance (s) = 18 m

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

• Deceleration (a)

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

→ Here we have to find the deceleration of the cyclist.

→ By the third equation of motion we know that,

  v² - u² = 2as

→ Substituting the given datas we get,

  0² - 15² = 2 × a × 18

  -225 = 36 a

   a = -225/36

   a = -6.25 m/s²

→ Here acceleration is negative since it is deceleration or retardation.

→ Therefore, the deceleration of the cyclist is -6.25 m/s²

$\boxed{\bold{Deceleration=-6.25\:m/s^{2}}}$

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

→ The three equations of motion are:

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