Question

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.​

Answer 1

Question 1 :-

Using the second equation of motion .

♦ v² - u² = 2as

Where ,

• v → final velocity = 0 m/s . Because the bus is at rest after motion .
• u → initial velocity = 30 m/s
• a → acceleration = ( v - u )/t
• t → time = ?

☞ 0² - ( 30 )² = 2 [ ( 0 - 30 )/t ] 50

☞ 0 - 900 = 100 [ -30/t ]

☞ - 900 = -3000/t

☞ t = -3000/-900

☞ t = 30/9 = 10/3

☞ Time = 3.33 s

Hence , time taken by the bus to stop after applying brakes = 3.33s

Question 2 :-

Using third equation of motion

♦ s = ut + 1/2 at²

Substituting the know values ,

☞ s = (0)(5) + 1/2 (10)(5)²

Initial velocity ( u ) will be zero because , the body has started from rest

☞ s = 0 + 1/2 (10) (25)

☞ s = 5 × 25

☞ Height of the building (s) = 125 m

Hence, height of the building = 125 m

Answer 2

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 :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ Second Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2as}}}$

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,

$\implies \sf 0 =\: 30 + at$

$\implies \sf 0 - 30 =\: at$

$\implies \sf \dfrac{0 - 30}{t} =\: a$

$\implies \sf\bold{\purple{a =\: \dfrac{- 30}{t}\: m/s^2}}$

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,

$\longrightarrow \sf (0)^2 =\: (30)^2 + 2 \times \bigg(\dfrac{- 30}{t}\bigg) \times 50$

$\longrightarrow \sf 0 =\: 900 + \dfrac{- 60}{t} \times 50$

$\longrightarrow \sf 0 =\: 900 + \dfrac{- 3000}{t}$

$\longrightarrow \sf 0 - 900 =\: \dfrac{- 3000}{t}$

$\longrightarrow \sf - 900 =\: \dfrac{- 3000}{t}$

By doing cross multiplication we get,

$\longrightarrow \sf - 900t =\: - 3000$

$\longrightarrow \sf t =\: \dfrac{\cancel{-} 3000}{\cancel{-} 900}$

$\longrightarrow \sf t =\: \dfrac{30\cancel{00}}{9\cancel{00}}$

$\longrightarrow \sf t =\: \dfrac{30}{9}$

$\longrightarrow \sf\bold{\red{t =\: 3.33\: seconds}}$

$\therefore$ 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 :

$\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}$

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,

$\longrightarrow \sf s =\: (0)(5) + \dfrac{1}{2} \times 10 \times (5)^2$

$\longrightarrow \sf s =\: 0 \times 5 + \dfrac{1}{2} \times 10 \times 5 \times 5$

$\longrightarrow \sf s =\: 0 + \dfrac{1}{2} \times 10 \times 25$

$\longrightarrow \sf s =\: 0 + \dfrac{1}{2} \times 250$

$\longrightarrow \sf s =\: 0 + \dfrac{250}{2}$

$\longrightarrow \sf s =\: \dfrac{0 + 250}{2}$

$\longrightarrow \sf s =\: \dfrac{250}{2}$

$\longrightarrow \sf\bold{\red{s =\: 125\: m}}$

$\therefore$ The height of the building is 125 m .