Q1. A bus is moving at 30 m/s. Driver decreased its speed and bus stopped after moving 50m. Find time required to stop the bus after applying brakes. Q2. A stone is gently released from top of the building. It strikes the ground in 5 second. Find height of the building. Take value of acceleration as 10 m/s/s.
Answers
Answer:
Answer:
Question No 1 :-
A bus is moving at 30 m/s. Driver decreased its speed and bus stopped after moving 50 m. Find the time required to stop the bus after applying brakes.
Given :-
A bus is moving at 30 m/s. Driver decreased its speed and bus stopped after moving 50 m/s.
To Find :-
What is the time required to stop the bus after applying brakes.
Formula Used :-
\clubsuit♣ First Equation Of Motion Formula :
\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}↦
v=u+at
\clubsuit♣ Second Equation Of Motion Formula :
\begin{gathered}\mapsto \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2as}}}\\\end{gathered}
↦
v
2
=u
2
+2as
where,
v = Final Velocity
u = Initial Velocity
a = Acceleration
t = Time
s = Distance Covered
Solution :-
First, we have to find the value of acceleration :
Given :
Final Velocity = 0 m/s
Initial Velocity = 30 m/s
According to the question by using the formula we get,
\implies \sf 0 =\: 30 + at⟹0=30+at
\implies \sf 0 - 30 =\: at⟹0−30=at
\implies \sf \dfrac{0 - 30}{t} =\: a⟹
t
0−30
=a
\implies \sf\bold{\purple{a =\: \dfrac{- 30}{t}\: m/s^2}}⟹a=
t
−30
m/s
2
Now, we have to find the time required to stop the bus after applying brakes :
Given :
Final Velocity = 0 m/s
Initial Velocity = 30 m/s
Acceleration = - 30/t m/s²
Distance Covered = 50 m
According to the question by using the formula we get,
\longrightarrow \sf (0)^2 =\: (30)^2 + 2 \times \bigg(\dfrac{- 30}{t}\bigg) \times 50⟶(0)
2
=(30)
2
+2×(
t
−30
)×50
\longrightarrow \sf 0 =\: 900 + \dfrac{- 60}{t} \times 50⟶0=900+
t
−60
×50
\longrightarrow \sf 0 =\: 900 + \dfrac{- 3000}{t}⟶0=900+
t
−3000
\longrightarrow \sf 0 - 900 =\: \dfrac{- 3000}{t}⟶0−900=
t
−3000
\longrightarrow \sf - 900 =\: \dfrac{- 3000}{t}⟶−900=
t
−3000
By doing cross multiplication we get,
\longrightarrow \sf - 900t =\: - 3000⟶−900t=−3000
\longrightarrow \sf t =\: \dfrac{\cancel{-} 3000}{\cancel{-} 900}⟶t=
−
900
−
3000
\longrightarrow \sf t =\: \dfrac{30\cancel{00}}{9\cancel{00}}⟶t=
9
00
30
00
\longrightarrow \sf t =\: \dfrac{30}{9}⟶t=
9
30
\longrightarrow \sf\bold{\red{t =\: 3.33\: seconds}}⟶t=3.33seconds
\therefore∴ The time required to stop the bus after applying brakes is 3.33 seconds.
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Question No 2 :-
A stone is gently released from top of the building. It strikes the ground in 5 second. Find the height of the building. (Take acceleration (a) = 10 m/s²).
Given :-
A stone is gently released from top of the building. It strikes the ground in 5 second. (a = 10 m/s²)
To Find :-
What is the height of the building.
Formula Used :-
\clubsuit♣ Third Equation Of Motion Formula :
\begin{gathered}\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}\\\end{gathered}
↦
s=ut+
2
1
at
2
where,
s = Distance Covered or distance
u = Initial Velocity
t = Time
a = Acceleration
Solution :-
Given :
Initial Velocity = 0 m/s
Time = 5 seconds
Acceleration = 10 m/s²
According to the question by using the formula we get,
\longrightarrow \sf s =\: (0)(5) + \dfrac{1}{2} \times 10 \times (5)^2⟶s=(0)(5)+
2
1
×10×(5)
2
\longrightarrow \sf s =\: 0 \times 5 + \dfrac{1}{2} \times 10 \times 5 \times 5⟶s=0×5+
2
1
×10×5×5
\longrightarrow \sf s =\: 0 + \dfrac{1}{2} \times 10 \times 25⟶s=0+
2
1
×10×25
\longrightarrow \sf s =\: 0 + \dfrac{1}{2} \times 250⟶s=0+
2
1
×250
\longrightarrow \sf s =\: 0 + \dfrac{250}{2}⟶s=0+
2
250
\longrightarrow \sf s =\: \dfrac{0 + 250}{2}⟶s=
2
0+250
\longrightarrow \sf s =\: \dfrac{250}{2}⟶s=
2
250
\longrightarrow \sf\bold{\red{s =\: 125\: m}}⟶s=125m
\therefore∴ The height of the building is 125 m .
Explanation:
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Answer:
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