Question

A bus running at a speed 18km/h is stop in 2.5 sec by applying break calculate retardation

Answer 1

ANSWER :

✴ Given :

▪ Initial velocity of bus = 18kmph

▪ Final velocity of bus = zero

▪ Time interval = 2.5s

✴ To Find :

▪ Retardation of bus

✴ Concept :

✏ Acceleration is defined as the ratio of change in velocity to the time interval.

✏ It is a vector quantity.

✏ It can be positive, negative and zero.

✴ Conversion :

✒ 1kmph = 5/18mps

✒ 18kmph = 18×5/18 = 5mps

✴ Calculation :

$\implies\rm\:a=\dfrac{v-u}{t}\\ \\ \implies\rm\:a=\dfrac{0-5}{2.5}\\ \\ \implies\rm\:a=-\dfrac{5}{2.5}\\ \\ \implies\boxed{\bf{\purple{a=-2\:ms^{-2}}}}$

⏭ Negative sign shows retardation.

Answer 2

Explanation:

$\bf \huge \: \: Question\: \:$

A bus running at a speed 18km/h is stop in 2.5 sec by applying break calculate retardation

_____________________________

$\bf \huge \: \: To\: Find \:$

• applying break calculate retardation
• a = ????

____________________________

$\bf \huge \: \: Given\:$

• A bus running at a speed 18km/h
• stop in 2.5 sec

_____________________________

We know The terms:-

• U= Initial speed
• V = Final Speed
• A =. Acceleration
• R = Retardation
• T = Time

___________________________

Initial speed of bus = U = 18 km/h

$\bf \:</strong><strong>=</strong><strong>5</strong><strong>/</strong><strong>1</strong><strong>8</strong><strong>×</strong><strong>1</strong><strong>8</strong><strong>=</strong><strong>5</strong><strong>\:$

$\bf \:</strong><strong>T</strong><strong> </strong><strong>=</strong><strong> </strong><strong>2.5 sec</strong><strong>\:$

We Have Formula :-

$\bf \:</strong><strong> </strong><strong> </strong><strong>V</strong> = u + at\:$

Putting all values:-

$\bf \:</u></strong><strong><u>=</u></strong><strong><u>></u></strong><strong><u> </u></strong><strong><u>0</u></strong><strong><u> </u></strong><strong><u>=</u></strong><strong><u> </u></strong><strong><u>5</u></strong><strong><u> </u></strong><strong><u>+</u></strong><strong><u>a</u></strong><strong><u>(</u></strong><strong><u>2</u></strong><strong><u>.</u></strong><strong><u>5</u></strong><strong><u>)</u></strong><strong><u>\:$

$\bf \:</u></strong><strong><u>=</u></strong><strong><u>></u></strong><strong><u>2</u></strong><strong><u>.</u></strong><strong><u>5</u></strong><strong><u>a</u></strong><strong><u> </u></strong><strong><u>=</u></strong><strong><u> </u></strong><strong><u>5</u></strong><strong><u>\:$

$\bf \:</u></strong><strong><u> </u></strong><strong><u>=</u></strong><strong><u>></u></strong><strong><u> </u></strong><strong><u>A</u></strong><strong><u> </u></strong><strong><u>=</u></strong><strong><u> </u></strong><strong><u>5</u></strong><strong><u>/</u></strong><strong><u>2</u></strong><strong><u>.</u></strong><strong><u>5</u></strong><strong><u>\:$

$\bf \:</u></strong><strong><u>=</u></strong><strong><u> </u></strong><strong><u>></u></strong><strong><u> </u></strong><strong><u>A</u></strong><strong><u> </u></strong><strong><u>=</u></strong><strong><u> </u></strong><strong><u>2</u></strong><strong><u>.</u></strong><strong><u>5</u></strong><strong><u> </u></strong><strong><u>m</u></strong><strong><u>/</u></strong><strong><u>s</u></strong><strong><u>^</u></strong><strong><u>2</u></strong><strong><u>\:$

$\bf \red{Hence. A = 2.5 m/s^2}\:$

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