Question

A car having mass of 1500 kg is moving at 60 km per hour when brakes are applied to produce a constant acceleration of the car stops in 1.2 minutes determine the braking force applied

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A car having mass of 1500 kg is moving at 60 km per hour when brakes are applied to produce a constant acceleration of the car stops in 1.2 minutes determine the Breaking force applied?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of the Body (M) = 1500 Kg.
• Initial velocity of the Car (u) = 60 km/h.
• Final velocity of the Car (v) = 0 km/h.

$\huge{\bold{\underline{Explanation:-}}}$

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Converting, The Given Units into S.I units,

Velocity:-

$\large{\tt u = 60 km/h}$

$\large{\tt \leadsto u = 60 \times \dfrac{5}{18}}$

$\large{\tt \leadsto u = \cancel{60} \times \dfrac{5}{\cancel{18}}}$

$\large{\tt \leadsto u = 10 \times \dfrac{5}{3}}$

$\large{\tt \leadsto u = \dfrac{50}{3}}$

$\large{\leadsto {\underline{\underline{\tt u = 60 \: Km/h = \dfrac{50}{3} \: m/s}}}}$

Time:-

$\large{\tt t = 1.2 \: minutes}$

$\large{\tt \leadsto t = 1.2 \times 60 }$

$\large{\tt \leadsto t = 12 \times 6}$

$\large{\tt \leadsto t = 72 }$

$\large{\leadsto {\underline{\underline{\tt t = 1.2 \: min = 72 \: Sec}}}}$

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Applying First kinematics equation,

$\large{\boxed{\tt v = u + at}}$

Substituting the values,

$\large{\tt \leadsto 0 = \dfrac{50}{3} + a \times 72}$

$\large{\tt \leadsto 0 = \dfrac{50}{3} + 72a}$

$\large{\tt \leadsto 72a = \dfrac{- 50}{3}}$

$\large{\tt \leadsto a = \dfrac{- 50}{3 \times 72}}$

$\large{\tt \leadsto a = \dfrac{-50}{216}}$

$\large{\tt \leadsto a = \cancel{\dfrac{-50}{216}}}$

$\large{\boxed{\tt a = - 0.23 \: m/s^2}}$

So, the deceleration of the car is 0.23 m/s².

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Applying Newton's second law of motion,

$\large{\boxed{\tt F = ma}}$

Substituting the values,

$\large{\tt \leadsto F = 1500 \times - 0.23}$

$\large{\tt \leadsto F = 15 \times - 23}$

$\huge{\boxed{\boxed{\tt F = - 345 \: N }}}$

So, the Breaking Force applied to stop the car is - 345 Newton's.

Note:-

• Here the Negative value of Force depicts that Force Applied is opposing the motion of the car.

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