Question

a bus is moving with a velocity of 72km/h then driver applies brake to stop the bus in 20s .then distance covered by bus in given time .
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Answer 1

• Initial velocity of the bus, u = 72 km/hr

$1 \: \frac{km}{hr} = \frac{5}{18} \: \frac{m}{s} \\ \\ 72 \: \frac{km}{hr} \: = 72 \times \frac{5}{18} \: \dfrac{m}{s} \\ \\ = \dfrac{72}{18} \times 5 \: \frac{m}{s} \\ \\ = 4 \times 5 \: \frac{m}{s} \\ \\ u = 20 \: \frac{m}{s}$

• Time after which the bus stops by the application of brake, t = 20 s

• Since the bus stops due to application of brake, the final velocity of the bus is zero. v = 0

• Let the distance covered by the bus be s.

According to the first equation of motion,

v = u + at

Where a is the negative acceleration or retardation.

$v = u + at \\ \\ 0 = 20 +( a \times 20) \\ \\ - 20 = 20a \\ \\ a = \frac{ - 20}{20} \\ \\ a = - 1 \: \dfrac{m}{ {s}^{2} }$

The retardation is -1 m/s².

According to third equation of motion,

${ v}^{2} - {u}^{2} = 2as \\ \\ {0}^{2} - {(20)}^{2} = 2 \times ( - 1) \times s \\ \\ - 400 = - 2s \\ \\ s = \dfrac{ - 400}{ - 2} \\ \\ s = 200 \: m$

The distance covered by the bus in the given time is 200 m.

Answer 2

Answer:

Given :-

• A bus is moving with a velocity of 72 km/h then driver applies brake to stop the bus in 20 seconds.

To Find :-

• What is the distance covered by bus.

Formula Required :-

✿ a = v - u/t

✿ v² - u² = 2as

Solution :-

Given :-

⇒ 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

Hence, acceleration will be - 1 m/s²

Again,

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.