Question

M.G.O.<br />A bus increases its speed From 36 km/h to 72 km/h<br />in losec it sseleration is :-<br />a) 5 m/ b) 2 m/s C) 3.6m/ d) Im/s.​

Answer 1

Correct Question :-

A bus increases its speed From 36 km/h to 72 km/h in 10 sec . Then find it's Acceleration.

To Find :-

The Acceleration of the bus.

Given :-

• Initial Velocity (u) = 36 km/h

• Final Velocity (v) = 72 km/h

• Time Taken (t) = 10 s

We Know :-

First Equation of Motion –

$\large{\boxed{\bf{v = u + at}}}$

Where :-

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration due to Gravity
• t = Time Taken

Concept :-

• The Acceleration Produced will be Positive as the bus in increasing it's speed from the initial point.

• Derivation of First Equation of Motion :-

To Derive :-

v = u + at

Consider a body having Initial Velocity u and Final Velocity after time t be v.

Hence , we get :-

• Initial Velocity = u
• Final Velocity = v
• Time Taken = t

Rate of change of Velocity = $\bf{\dfrac{v - u}{t}}$

Let the Rate of change of Velocity be a.

Hence, we get the equation as :-

$:\implies \bf{a = \dfrac{v - u}{t}}$

By solving the above equation , we get :-

$:\implies \bf{a \times t = v - u} \\ \\ \\ :\implies \bf{at = v - u} \\ \\ \\ :\implies \bf{at + u = v} \\ \\ \\ \therefore \purple{\bf{v = u + at}}$

Thus , the equation formed is v = u + at !!

Solution :-

• u = 36 km/h
• v = 72 km/h
• t = 10 s

First to have to make the same in each case , so let's convert Initial Velocity and Final Velocity into m/s from km/hr.

To convert a unit from km/h to m/s , multiply it 5/18.i.e

• u = 36 km/h

➝ $\bf{\bigg(36 \times \dfrac{5}{18}\bigg)ms^{-1}}$

➝ $\bf{\bigg(2 \times \dfrac{5}{1}\bigg)ms^{-1}}$

➝ $\bf{10 ms^{-1}}$

Hence, the Initial Velocity in m/s is 10 m/s.

• v = 72 km/h

➝ $\bf{\bigg(72 \times \dfrac{5}{18}\bigg)ms^{-1}}$

➝ $\bf{\bigg(4 \times \dfrac{5}{1}\bigg)ms^{-1}}$

➝ $\bf{20 ms^{-1}}$

Hence, the Final Velocity in m/s is 20 m/s.

Acceleration Produced :-

• v = 20 m/s

• u = 10 m/s

• t = 10 s

Let the acceleration Produced be a.

Using the First Equation of Motion and substituting the values in it , we get :-

$:\implies \bf{v = u + at} \\ \\ \\ :\implies \bf{20 = 10 + a \times 10} \\ \\ \\ :\implies \bf{20 - 10 = 10a} \\ \\ \\ :\implies \bf{10 = 10a} \\ \\ \\ :\implies \bf{\dfrac{10}{10} = a} \\ \\ \\ :\implies \bf{1 = a} \\ \\ \\ \therefore \purple{\bf{a = 1 ms^{-2}}}$

Hence, the Acceleration Produced is 1 m/s².

Answer 2

Answer:

Correct Question :-

A bus increases its speed From 36 km/h to 72 km/h in 10 sec . Then find it's Acceleration.

To Find :-

The Acceleration of the bus.

Given :-

Initial Velocity (u) = 36 km/h

Final Velocity (v) = 72 km/h

Time Taken (t) = 10 s

We Know :-

First Equation of Motion –

\large{\boxed{\bf{v = u + at}}}

v=u+at

Where :-

v = Final Velocity

u = Initial Velocity

a = Acceleration due to Gravity

t = Time Taken

Concept :-

The Acceleration Produced will be Positive as the bus in increasing it's speed from the initial point.

Derivation of First Equation of Motion :-

To Derive :-

v = u + at

Consider a body having Initial Velocity u and Final Velocity after time t be v.

Hence , we get :-

Initial Velocity = u

Final Velocity = v

Time Taken = t

Rate of change of Velocity = \bf{\dfrac{v - u}{t}}

t

v−u

Let the Rate of change of Velocity be a.

Hence, we get the equation as :-

:\implies \bf{a = \dfrac{v - u}{t}}:⟹a=

t

v−u

By solving the above equation , we get :-

\begin{gathered}:\implies \bf{a \times t = v - u} \\ \\ \\ :\implies \bf{at = v - u} \\ \\ \\ :\implies \bf{at + u = v} \\ \\ \\ \therefore \purple{\bf{v = u + at}}\end{gathered}

:⟹a×t=v−u

:⟹at=v−u

:⟹at+u=v

∴v=u+at

Thus , the equation formed is v = u + at !!

Solution :-

u = 36 km/h

v = 72 km/h

t = 10 s

First to have to make the same in each case , so let's convert Initial Velocity and Final Velocity into m/s from km/hr.

To convert a unit from km/h to m/s , multiply it 5/18.i.e

u = 36 km/h

➝ \bf{\bigg(36 \times \dfrac{5}{18}\bigg)ms^{-1}}(36×

18

5

)ms

−1

➝ \bf{\bigg(2 \times \dfrac{5}{1}\bigg)ms^{-1}}(2×

1

5

)ms

−1

➝ \bf{10 ms^{-1}}10ms

−1

Hence, the Initial Velocity in m/s is 10 m/s.

v = 72 km/h

➝ \bf{\bigg(72 \times \dfrac{5}{18}\bigg)ms^{-1}}(72×

18

5

)ms

−1

➝ \bf{\bigg(4 \times \dfrac{5}{1}\bigg)ms^{-1}}(4×

1

5

)ms

−1

➝ \bf{20 ms^{-1}}20ms

−1

Hence, the Final Velocity in m/s is 20 m/s.

Acceleration Produced :-

v = 20 m/s

u = 10 m/s

t = 10 s

Let the acceleration Produced be a.

Using the First Equation of Motion and substituting the values in it , we get :-

\begin{gathered}:\implies \bf{v = u + at} \\ \\ \\ :\implies \bf{20 = 10 + a \times 10} \\ \\ \\ :\implies \bf{20 - 10 = 10a} \\ \\ \\ :\implies \bf{10 = 10a} \\ \\ \\ :\implies \bf{\dfrac{10}{10} = a} \\ \\ \\ :\implies \bf{1 = a} \\ \\ \\ \therefore \purple{\bf{a = 1 ms^{-2}}}\end{gathered}

:⟹v=u+at

:⟹20=10+a×10

:⟹20−10=10a

:⟹10=10a

:⟹

10

10

=a

:⟹1=a

∴a=1ms

−2

Hence, the Acceleration Produced is 1 m/s².