A bus travelling with a velocity 36kmph is brought to rest by applying brakes it experiences a uniform retardation of 4m/s^2.before coming to rest how much distance does it travel
Answers
Answer:
12.5 m
Explanation:
initial velocity =36km/h
36×5/18=10m/s
u=10m/s v or final velocity =0m/s as it came to rest
retardation =u-v/t
4=10-0/t
t=2.5 seconds
distance =ut+1/2at²
10×2.5+1/2×-4×2.5²
25-12.5=12.5 m
Answer:
⇝ Given :-
A jeep travelling with a velocity of 108 km/h.
It is brought to rest by applying brakes.
It experiences a uniform retardation of 3 m/s²
⇝ To Find :-
Distance travelled by Jeep before coming to rest.
⇝ Solution :-
❒ Converting Initial Velocity in m/s ;
\begin{gathered}\rm Initial \: Velocity = 108 \: km/h \\ \end{gathered}
InitialVelocity=108km/h
\begin{gathered} = \rm\bigg(108 \times \frac{5}{18} \bigg)m/s \\ \end{gathered}
=(108×
18
5
)m/s
\begin{gathered} = \rm \bigg(\frac{540}{18} \bigg)m/s \\ \end{gathered}
=(
18
540
)m/s
\begin{gathered}:\longmapsto \red{ \rm Initial \: Velocity = 30 \: m/s } \\ \end{gathered}
:⟼InitialVelocity=30m/s
❒ Calculating Required Distance :
We Have,
Initial Velocity = u = 30 m/s
Final Velocity = v = 0 m/s
Acceleration = a = - 3m/s²
Let time taken by Jeep to stop be = t second.
★ We Have 1st Equation of Motion as :
\begin{gathered}\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ v = u + at }}} \\ \end{gathered}
★
v=u+at
⏩ Putting Values In 1st Equation :
\begin{gathered}:\longmapsto \rm 0 = 30 + ( - 3) \times t \\ \end{gathered}
:⟼0=30+(−3)×t
\begin{gathered}:\longmapsto \rm \cancel - 3t = \cancel - 30 \\ \end{gathered}
:⟼
−
3t=
−
30
\begin{gathered}:\longmapsto \rm t = \cancel \frac{30}{3} \\ \end{gathered}
:⟼t=
3
30
\purple{ \large :\longmapsto \underline {\boxed{{\bf t = 10 \: s} }}}:⟼
t=10s
★ We Have 2nd Equation of Motion as :
\begin{gathered}\large \bf \red \bigstar \: \: \orange{ \underbrace{ \underline{s=ut+\dfrac{1}{2}at^2}}} \\\end{gathered}
★
s=ut+
2
1
at
2
⏩ Putting Values In 2nd Equation :
\begin{gathered}:\longmapsto \rm s = 30 \times 10 + \frac{1}{2} \times ( - 3) \times {10}^{2} \\ \end{gathered}
:⟼s=30×10+
2
1
×(−3)×10
2
\begin{gathered}:\longmapsto \rm s = 300 - 150 \\ \end{gathered}
:⟼s=300−150
\purple{ \large :\longmapsto \underline {\boxed{{\bf s = 150 \: m} }}}:⟼
s=150m
Therefore,
\large\underline{\pink{\underline{\frak{\pmb{Required \: Distance = 150 \: m}}}}}
RequiredDistance=150m
RequiredDistance=150m