Question

The velocity of a Bus is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m.Compute the acceleration.

Answer 1

GIVEN :-

• Initial velocity ( u ) = 15 m/s.
• Final velocity ( v ) = 7 m/s.
• Distance covered ( s ) = 90 m.

TO FIND :-

• The Acceleration ( a ).

SOLUTION :-

As we know that the third equation of motion,

$\\ : \implies \displaystyle \sf \: v ^{2} - u ^{2} = 2as$

$: \implies \displaystyle \sf \: (7) ^{2} - (15) ^{2} = 2 \times a \times 90$

$: \implies \displaystyle \sf \: 49 - 225 = 180a$

$: \implies \displaystyle \sf \: - 176 = 180a$

$: \implies \displaystyle \sf \: a = \frac{ - 176}{180}$

$: \implies \underline{ \boxed{ \displaystyle \sf \: a \approx - 0.97 \: ms ^{ - 2} }}$

Hence the acceleration of the bus is -0.97 m/s².

[ Note :- -ve sign in the acceleration shows the retardation of the acceleration. ]

Answer 2

$\tt \huge {\pink {Given}}$

• Initial velocity ( u ) = 15 m/s.
• Final velocity ( v ) = 7 m/s.
• Distance covered ( s ) = 90 m.

$\tt \huge {\blue{To \: Find }}$

• Acceleration ( a ).

$\tt \huge {\orange{solution }}$

we know that ,

$\begin{gathered} \\ : \implies \displaystyle \sf \: v ^{2} - u ^{2} = 2as \end{gathered}$

So ,

$\begin{gathered} : \implies \displaystyle \sf \: (7) ^{2} - (15) ^{2} = 2 \times a \times 90 \end{gathered}$

$\begin{gathered} : \implies \displaystyle \sf \: 49 - 225 = 180a \end{gathered}$

$\begin{gathered} : \implies \displaystyle \sf \: - 176 = 180a \end{gathered}$

$\begin{gathered} : \implies \displaystyle \sf \: a = \frac{ - 176}{180} \end{gathered}$

$\begin{gathered} {\underline { \boxed{ \displaystyle \sf \: a \approx - 0.97 \: ms ^{ - 2} }}}\\ \end{gathered}$

$\tt {\green {So ,\: the \:acceleration \:of \:the\: bus\: is\ -0.97 m/s².}}$