Question

A car moving with a speed of 50 km/h can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/h, the minimum stopping distance is:
a. 12m
b. 18m
c. 24m
d. 6m

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Answer 1

Answer:

$\bigstar{\bold{Option\:C:\:24\:m}}$

Explanation:

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

• Car was moving with a speed of 50 km/hr(u₁) = 13 .9  m/s
• ​Distance travelled before coming to rest (s₁)= 6 m
• Car moves with a speed of 100 km/hr (u₂) = 27.8 m/s
• Final velocity (v) = 0 m/s

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

• Distance trvelled before coming to rest (s₂)

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

→ First we need to find out the acceleration of the body.

→ By the third equation of motion,

  v² - u² = 2as

→ Substituting the datas we get the value of a

  0² - (13.9)² = 2 × a × 6

   a = -193.21/12

   a = -16 m/s²

→ Acceleration is negative since it is retardation.

→ Now putting the value of a in third equation of motion with velocity u₂

  0² - (27.8)² = 2 × -16 × s

  s = 772.84/32

  s = 24.1 m ≈ 24 m

$\boxed{\bold{Distance\:travelled=24\:m}}$

→ Hence option C is correct.

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

→ The three equations of motion are:

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