a car moving with a speed of 50 km per hour can be stopped by brakes a

1.	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» 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 ━━━━━━━━━━━━━ Thanks

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»

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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