Question

A car travelling at 110 km/h decelerates for 4.5 seconds to a final velocity of 45 km/h. Calculate the deceleration of the car​

Answer 1

Answer:

Deacceleration is 4m/s

2

\large \dag† Step by step Explanation :-

❒ Converting Velocities in m/s :-

We Have,

\begin{gathered}\text{Initial Velocity= 110 km/h} \\\end{gathered}

Initial Velocity= 110 km/h

\begin{gathered}= \rm\bigg( 110\times \frac{5}{18}\bigg)second \\ \end{gathered}

=(110×

18

5

)second

\red{:\longmapsto \text{Time = 30.5 \:m/s}}:⟼Time = 30.5 m/s

Also,

\begin{gathered}\text{Final Velocity= 45 km/h} \\\end{gathered}

Final Velocity= 45 km/h

\begin{gathered}= \rm\bigg( 45\times \frac{5}{18}\bigg)second \\ \end{gathered}

=(45×

18

5

)second

\begin{gathered}\red{:\longmapsto \text{Time = 12.5 \:m/s}}\\\end{gathered}

:⟼Time = 12.5 m/s

Now Here we have :

Initial Velocity = u = 30.5 m/s

Final Velocity = v = 12.5 m/s

Time = t = 4.5 s

Let acceleration be = a m/s²

❒ We Have 1st Equation of Motion as :

\begin{gathered}\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{\blue{v = u + at}}}}\\\end{gathered}

★

v=u+at

⏩ Applying 1st Equation of Motion ;

\begin{gathered} \\ :\longmapsto \rm 12.5 = 30.5 + a \times 4.5 \\ \\ \end{gathered}

:⟼12.5=30.5+a×4.5

\begin{gathered}:\longmapsto \rm 12.5 - 30.5 = 4.5a \\ \\ \end{gathered}

:⟼12.5−30.5=4.5a

\begin{gathered}:\longmapsto \rm 4.5a = - 18 \\ \\ \end{gathered}

:⟼4.5a=−18

\begin{gathered}:\longmapsto \rm a = \frac{ - 18}{ \: \: 4.5} \\ \\ \end{gathered}

:⟼a=

4.5

−18

\begin{gathered}\purple{ \large :\longmapsto \underline {\boxed{{\bf a = -4} }}}\\\end{gathered}

:⟼

a=−4

⇝ Acclerration is - 4 m/s²