Question

a car came to rest when breaker was applied for 4 seconds to get a retardation of 3 m/s2 calculate how far the car would have travelled after applying the brake​

Answer 1

Given :

• Time Taken = 4 s

• Retardation = 3 m/s² or Acceleration = (-3) m/s².

To find :

The distance traveled by the car.

Solution :

Since , the car is retarding the final Velocity of the car will be 0 , i.e, v = 0.

Now ,

To find the distance covered by the car, first we need to find the velocity of the car before Retarding.

Using the first Equation of Motion and substituting the values in it , we get :

$\underline{:\implies \bf{v = u + at}}$

Where :-

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration due to gravity
• t = Time Taken

$:\implies \bf{0 = u + (-3) \times 4}$

$:\implies \bf{-u = (-3) \times 4}$

$:\implies \bf{-u = - 12}$

$:\implies \bf{\not{-}u = \not{-} 12}$

$:\implies \bf{u = 12}$

$\underline{\therefore \bf{Initial\:Velocity\:(u) = 12\:ms^{-1}}}$

Hence, the final Velocity of the car before coming to retardation is 12 m/s.

To find the distance traveled after applying the brake :

Using the third Equation of Motion and substituting the values in it , we get :-

$\underline{:\implies \bf{v^{2} = u^{2} + 2aS}}$

• Where :-

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration due to gravity
• S = Distance

$:\implies \bf{v^{2} = u^{2} + 2aS}$

$:\implies \bf{0^{2} = 12^{2} - 2 \times 3 \times S}$

$:\implies \bf{0 = 12^{2} - 2 \times 3 \times S}$

$:\implies \bf{0 = 144 - 6S}$

$:\implies \bf{- 144 = - 6S}$

$:\implies \bf{\not{-} 144 = \not{-} 6S}$

$:\implies \bf{144 = 6S}$

$:\implies \bf{\dfrac{144}{6} = S}$

$:\implies \bf{24 = S}$

$\underline{\therefore \bf{Distance\:(S) = 24\:m}}$

Hence, the distance covered after retardation is 24 m.

Answer 2

$\huge{\dag}\:{\underline{\boxed{\sf{\purple{ANSWER~~}}}}}</p><p></p><p>$

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24 m

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a = -3 m/s^2

t = 4 s

v = 0

v = u+at

-u = -3 ×4+0

u = 12 m/s

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Displacement of Car

$\implies \: s = ut + \frac{1}{2} at {}^{2}$

$\implies \: (12 \times 4) + \frac{1}{2} ( - 3) \times 16$

$\implies \: 24 \: m$