Question

A bus increases its speed from 72 km/h to180 km/h in 10 seconds. Its acceleration is?

a} 6 metre per second square

a} 5 metre per second square

a} 3 metre per second square

a} 8 metre per second square


correct uttar ko hi brainliest mark krungi -,-​

Answer 1

Given :

• A bus increases its speed from 72 km/h to180 km/h in 10 seconds.

To Find :

• Acceleration of the body .

Solution :

⟼ InitialVelocity = 72km/hr = 20m/s

⟼ FinalVelocity = 180km/hr = 50m/s

⟼ Time Taken = 10sec

Using Formula :

⟼Acceleration=t/v−u

Substituting Values :

⟼10/50−20

⟼10/30

⟼ 3 m/s−²

_____________________

Acceleration :

• It is defined as the rate of change of velocity of the object .

Formula of Acceleration is :-

⟶Acceleration = t/v−u

where,

• v = Final Velocity
• u = Initial Velocity
• t = time taken

_____________________

Answer 2

Provided that:

Initial velocity = 72 kmphFinal velocity = 180 kmphTime = 10 seconds

To calculate:

The acceleration

Solution:

The acceleration = 3 metre per second sq.

Using concepts:

Acceleration formulaFormula to convert kmph-mps

Using formulas:

Required solution:

~ Firstly let us convert kmph-mps

Converting 72 kmph into mps

Converting 180 kmph into mps

Therefore,

Initial velocity = 20 mpsFinal velocity = 50 mps

~ Now let's calculate the acceleration!

\begin{lgathered}:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{50-20}{10} \\ \\ :\implies \sf a \: = \dfrac{30}{10} \\ \\ :\implies \sf a \: = 3 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 3 \: ms^{-2}\end{lgathered}:⟹Acceleration=TimeChangeinvelocity:⟹a=tv−u:⟹a=1050−20:⟹a=1030:⟹a=3ms−2:⟹Acceleration=3ms−2

Henceforth, acceleration = 3 m/s sq. Therefore, option C is correct.

Additional information:

Something about acceleration!

\begin{lgathered}\begin{gathered}\boxed{\begin{array}{c}\\ \bf What \: is \: acceleration? \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. \\ \\ \sf \star \: Negative \: acceleration \: is \: known \: as \: deacceleration. \\ \sf \star \: Deacceleration \: is \: known \: as \: retardation. \\ \sf \star \: It's \: SI \: unit \: is \: ms^{-2} \: or \: m/s^2 \\ \sf \star \: It \: may \: be \: \pm ve \: or \: 0 \: too \\ \sf \star \: It \: is \: a \: vector \: quantity \\ \\ \bf Conditions \: of \pm ve \: or \: 0 \: acceleration \\ \\ \sf \odot \: Positive \: acceleration: \: \sf When \: \bf{u} \: \sf is \: lower \: than \: \bf{v} \\ \sf \odot \: Negative \: acceleration: \: \sf When \: \bf{v} \: \sf is \: lower \: than \: \bf{u} \\ \sf \odot \: Zero \: acceleration: \: \sf When \: \bf{v} \: \sf and \: \bf{u} \: \sf are \: equal \end{array}}\end{gathered}\end{lgathered}Whatisacceleration?Therateofchangeofvelocityofanobjectwithrespecttotimeisknownasacceleration.⋆Negativeaccelerationisknownasdeacceleration.⋆Deaccelerationisknownasretardation.⋆It′sSIunitisms−2orm/s2⋆Itmaybe±veor0too⋆ItisavectorquantityConditionsof±veor0acceleration⊙Positiveacceleration:Whenuislowerthanv⊙Negativeacceleration:Whenvislowerthanu⊙Zeroacceleration:Whenvanduareequal