Question

A bus is moving with a velocity of 72 km/h then driver applies brake to stop the bus in 20 sec. then distance covered by bus in given time.

-100 m

100 m

200 m

300 m ​

Answer 1

Given :

Initial velocity of bus = 72km/hr

Final velocity = zero

Time interval = 20s

To Find :

Distance covered by bus before coming to rest.

Solution :

❖ First of all we need to find acceleration of the bus. Acceleration is defined as the rate of change of velocity$.$

• It is a vector quantity having both magnitude as well as direction.
• It can be positive, negative or zero.
• SI unit : m/s²

Formula : a = (v - u) / t

» v denotes final velocity

» u denotes initial velocity

» a denotes acceleration

» t denotes time

By substituting the given values;

• u = 72km/hr = 20m/s

➙ a = (0 - 20) / 20

➙ a = -20/20

➙ a = -1 m/s²

[Negative sign shows retardation]

Applying 3rd equation of kinematics;

➠ v² - u² = 2as

➠ 0² - 20² = 2(-1)s

➠ -400 = -2s

➠ s = 400/2

➠ s = 200 m

Answer 2

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$\star \: \bf\underbrace{Question} \: \star$

• A bus is moving with a velocity of 72 km/h then driver applies brake to stop the bus in 20 sec. then distance covered by bus in given time___?

$\text{\large\underline{\purple{Given:- }}}$

• Initial velocity of bus = 72km/hr
• Final velocity = zero
• Time interval = 20s

$\text{\large\underline{\pink{To Find }}}$

• Distance covered by bus before coming to rest.

Formula Required :-

• a = v - u/t
• v² - u² = 2as

$\text{\large\underline{\orange{Solution:- }}}$

• ⇒ v = 0 m/s
• ⇒ u = 72 km/h = 72 ×$\dfrac{5}{18}$m/s = 20 m/s
• t = 20 s

To find acceleration,

• a =$\dfrac{0 - 20}{20}$

• a = $\dfrac{- 20}{20}$
• a = - 1 m/s²

$\text{\large\underline{\red{Hence,}}}$

• Acceleration will be - 1 m/s²

To find distance covered,

• (0)² - (20)² = 2 × (- 1) × s

• 0 - 400 = 2 × (- 1) × s

• - 400 = - 2s

• $\dfrac{{\cancel{- 400}}}{{\cancel{- 2}}}$= S

• s = 200 m

Therefore,

the distance covered by bus is 200 m.

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