Question

a car moving with a speed of 50 km per hour can be stopped by brakes at least 6 m what is the minimum stopping distance of the car is moving at a speed of 100 km per hour​

Answer 1

Answer

Given -

u = 50 km/hr

v = 0 km/hr ( because car is stopped)

s = 6 m

where

$\implies$u is initial velocity

$\implies$v is final velocity

$\implies$s is distance travelled.

━━━━━━━━━━━━━

To find -

Distance when initial velocity is 100 km

━━━━━━━━━━━━━

Formula used -

Third equation of motion -

$\implies$v² = u² + 2as

━━━━━━━━━━━━━

Solution -

$\longrightarrow$u = 50 km/hr

$\longrightarrow$v = 0 km/hr

$\longrightarrow$s = 6m

Substituting the value -

v² = u² + 2as

0 = (50)² + 2 × a × 6

a = -58 m/s²

━━━━━━━━━━━━━

$\longrightarrow$u = 100 km/hr = 27.7 m/s

$\longrightarrow$v = 0

$\longrightarrow$a = -58 m/s²

Substituting the value

v² = u² + 2as

0 = (27.7)² - 2 × 58 × s

s = 6.6 m

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Thanks

Answer 2

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Given -

u = 50 km/hr

v = 0 km/hr ( because car is stopped)

s = 6 m

where

⟹ u is initial velocity

⟹ v is final velocity

⟹ s is distance travelled.

━━━━━━━━━━━━━

To find -

Distance when initial velocity is 100 km

━━━━━━━━━━━━━

Formula used -

Third equation of motion -

⟹ v² = u² + 2as

━━━━━━━━━━━━━

Solution -

⟶ u = 50 km/hr

⟶ v = 0 km/hr

⟶ s = 6m

Substituting the value -

v² = u² + 2as

0 = (50)² + 2 × a × 6

a = -58 m/s²

━━━━━━━━━━━━━

⟶ u = 100 km/hr = 27.7 m/s

⟶ v = 0

⟶ a = -58 m/s²

Substituting the value

v² = u² + 2as

0 = (27.7)² - 2 × 58 × s

s = 6.6 m

━━━━━━━━━━━━━