Question

a car is travelling at a speed of 30km/h brakes are applied as to produce an uniform acceleration of - 0.5 m/s^2. Find how far the car will go before it is brought to rest

Answer 1

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✤ Required Answer:

✒ GiveN:

• Initially, Speed of the car = 30 km/h
• Accelerated at the rate of -0.5 m/s²
• Finally, It comes to rest.

✒ To FinD:

• Distance travelled by car before it stops$.$

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✤ How to solve?

For solving the question, we need to know the three equations of motion, that are:

• v = u + at
• s = ut + 1/2 at²
• v² = u² + 2as

So, Here:

• u = Initial velocity
• v = Final velocity
• s = Distance travelled
• t = Time taken
• a = Acceleration

⚡So, let's solve this question....

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✤ Solution:

We have,

• u = 30 km/h

➝ 1 km/h = 5/18 m/s

➝ 30 km/h = 30 × 5/18 m/s

➝ 30 km/h = 25/3 m/s

• a = - 0.5 m/s²
• v = 0 m/s [It comes to rest]

By using 3rd equation of motion,

➝ v² = u² + 2as

➝ 0² = (25/3)² + 2(-0.5)s

➝ 0 = 625/9 - s

Flipping to other side,

➝ 625/9 - s = 0

➝ s = 625/9 m

➝ s = 69.45 m(approx.)

⚘ Distance travelled by the car = 69.45 m

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Answer 2

$\large\bf{\underbrace{\red{SOLUTION:-}}}$

GIVEN :-

• car is traveling at a speed of 30km/h .
• Brakes are applied as to produce an uniform acceleration of -0.5m/s² .

• Initial Velocity (u) = 30 km/h = 25/3 m/s
• Retardation (a) = -0.5 m/s²

☞ As breaks are applied, so the final velocity of the car is ‘0 m/s’ .

• Final velocity (v) = 0 m/s

FORMULA :-

✴️ When a body accelerates uniformly, then the Equation of Motions are,

$\bigstar\:\rm{\purple{\boxed{v\:=\:u\:+\:at\:}}}$

$\bigstar\:\rm{\blue{\boxed{S\:=\:ut\:+\:\dfrac{1}{2}\:at^2\:}}}$

$\bigstar\:\rm{\pink{\boxed{v^2\:=\:u^2\:+\:2as\:}}}$

☞ Here,

• u = initial velocity
• v = final velocity
• a = acceleration or retardation
• t = time
• s = Distance

To Find :-

• The distance covered by the car, before it brought to rest .

CALCULATION :-

$\rm{\implies\:0^2\:-\:(\dfrac{25}{3})^2\:=\:2\times{-0.5}\times{s}\:}$

$\rm{\implies\:-\dfrac{625}{9}\:=\:-1\times{s}\:}$

$\rm{\implies\:s\:=\:\dfrac{625}{9}\:m}$

$\rm{\boxed{\implies\:s\:=\:69.44m\:}}$

✴️ Therefore, the car will go ‘69.44m’ before it is brought to rest .