Question

a bus is moving along a straight road with a speed of 126 km per hour is brought to stop within a distance of 500 m. what is retardation of bus and how long does it take for bus to stop
please give me full Answer​

Answer 1

Answer:

Retardation = + 1.255 m/s²

Time = 28.57 s

Explanation:

As per the provided information in the given question, it has been provided to us that a bus's initial velocity is 126 km/h and then it comes to rest after covering the distance of 500 m. We've been asked to calculate retardation and time taken. We have,

• Initial velocity (u) = 126 km/h
• Final velocity (v) = 0 m/s (As it comes to rest)
• Distance covered (s) = 500 m

Before commencing the steps, firstly we need to convert initial velocity in m/s.

$\longrightarrow \sf{\quad { u = 126 \; km \; h^{-1}}}$

• 1 km/h = 5/18 m/s

$\longrightarrow \sf{\quad { u =\Bigg \lgroup \cancel{126} \times \dfrac{5}{\cancel{18}} \Bigg \rgroup \; m \; s^{-1}}}$

$\longrightarrow \sf{\quad { u =\Bigg \lgroup 7 \times 5 \Bigg \rgroup \; m \; s^{-1}}}$

$\longrightarrow \quad \boxed{\sf{ u = 35 \; m \; s^{-1} } }$

Now, we'll calculate the acceleration first to calculate the retardation. By using the third equation of motion,

$\longrightarrow \quad \boxed{\textbf{\textsf{v}}^\textbf{\textsf{2}} - \textbf{\textsf{u}}^\textbf{\textsf{2}} = \textbf{\textsf{2as}} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

$\longrightarrow \sf{\quad {(0)^2 - (35)^2 =2(a)(500) }}$

$\longrightarrow \sf{\quad {0 - 1225 = 1000a}}$

$\longrightarrow \sf{\quad {- 1225 = 1000a}}$

$\longrightarrow \sf{\quad { \cancel{\dfrac{ - 1225}{1000}} = a}}$

$\longrightarrow \sf{\quad {- 1.225 =a }}$

Retardation is nothing but the acceleration with negative sign so,

$\longrightarrow \quad \underline{\boxed {\textbf{\textsf{ +1.225}} \; \textbf{\textsf{m}} \; \textbf{\textsf{s}}^{\textbf{\textsf{-2}}} = \textbf{\textsf{Retardation}} }}$

$\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

Now, we have to calculate time taken. By using the first equation of motion,

$\longrightarrow \quad \boxed{\textbf{\textsf{v = u + at}}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\longrightarrow \sf{\quad { 0 = 35 + (- 1.225)a}}$

$\longrightarrow \sf{\quad { 0 - 35 =- 1.225t}}$

$\longrightarrow \sf{\quad { - 35 =- 1.225t}}$

$\longrightarrow \sf{\quad { \cancel{\dfrac{ - 35 }{-1.225}}= t}}$

$\longrightarrow \quad \underline{\boxed{ \textbf{\textsf{ 28.57 \; s = t}}}}$

$\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

Therefore,

⠀⠀⠀⠀★ Retardation = + 1.255 m/s²

⠀⠀⠀⠀★ Time = 28.57 s