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​

Answer 1

Provided that:

• Initial velocity = 72 kmph
• Final velocity = 180 kmph
• Time = 10 seconds

To calculate:

• The acceleration

Solution:

• The acceleration = 3 metre per second sq.

Using concepts:

• Acceleration formula
• Formula to convert kmph-mps

Using formulas:

• ${\small{\underline{\boxed{\pmb{\sf{1 \: kmph \: = \dfrac{5}{18} \: mps}}}}}}$

• ${\small{\underline{\boxed{\pmb{\sf{a \: = \dfrac{v-u}{t}}}}}}}$

Required solution:

~ Firstly let us convert kmph-mps

Converting 72 kmph into mps

$:\implies \sf 72 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{72} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 4 \times 5 \\ \\ :\implies \sf 20 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Converting 180 kmph into mps

$:\implies \sf 180 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{180} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 10 \times 5 \\ \\ :\implies \sf 50 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Therefore,

• Initial velocity = 20 mps
• Final velocity = 50 mps

~ Now let's calculate the acceleration!

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{50-20}{10} \\ \\ :\implies \sf a \: = \dfrac{30}{10} \\ \\ :\implies \sf a \: = 3 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 3 \: ms^{-2}$

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

Additional information:

Something about acceleration!

$\begin{gathered}\boxed{\begin{array}{c}\\ \bf What \: is \: acceleration? \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. \\ \\ \sf \star \: Negative \: acceleration \: is \: known \: as \: deacceleration. \\ \sf \star \: Deacceleration \: is \: known \: as \: retardation. \\ \sf \star \: It's \: SI \: unit \: is \: ms^{-2} \: or \: m/s^2 \\ \sf \star \: It \: may \: be \: \pm ve \: or \: 0 \: too \\ \sf \star \: It \: is \: a \: vector \: quantity \\ \\ \bf Conditions \: of \pm ve \: or \: 0 \: acceleration \\ \\ \sf \odot \: Positive \: acceleration: \: \sf When \: \bf{u} \: \sf is \: lower \: than \: \bf{v} \\ \sf \odot \: Negative \: acceleration: \: \sf When \: \bf{v} \: \sf is \: lower \: than \: \bf{u} \\ \sf \odot \: Zero \: acceleration: \: \sf When \: \bf{v} \: \sf and \: \bf{u} \: \sf are \: equal \end{array}}\end{gathered}$

Answer 2

Given :

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

To Find :

• Acceleration of the body .

Solution :

$\longmapsto\tt{Initial\:Velocity=72\:km/hr=20\:m/s}$

$\longmapsto\tt{Final\:Velocity=180\:km/hr=50\:m/s}$

$\longmapsto\tt{Time\:Taken=10\:sec}$

Using Formula :

$\longmapsto\tt\boxed{Acceleration=\dfrac{v-u}{t}}$

Putting Values :

$\longmapsto\tt{\dfrac{50-20}{10}}$

$\longmapsto\tt{\dfrac{30}{10}}$

$\longmapsto\tt\bf{3\:{m/s}^{-2}}$

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Acceleration :

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

Formula of Acceleration is :-

$\longrightarrow\tt{Acceleration=\dfrac{v-u}{t}}$

Here :

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

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