Question

1) The velocity of a body changes from 65 m/s - 98 m/s by Time of 12 s . What is distance covered by body if acceleration is 2.75 m/s².

2) A body starts from rest and acquires a velocity of 18 km/h in 10 s. The acceleration of the body is ?

3) a body of mass 10 kg is moving with a velocity of 5 m/s attains of 35m/s in 5 second find the force applied on body to change its velocity ?

4) A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.
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Answer 1

Answer:

✪ Solution No 1 :- ✪

Given :

• The velocity of a body changes from 65 m/s to 98 m/s by time of 12 seconds.
• The acceleration is 2.75 m/s².

To Find :-

• What is the distance covered by the body.

Solution :-

Given :

• Initial Velocity = 65 m/s
• Final Velocity = 98 m/s
• Time Taken = 12 seconds
• Acceleration = 2.75 m/s²

As we know that :

$\clubsuit$ Second Equation Of Motion Formula :

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

where,

• s = Distance Covered
• u = Initial Velocity
• t = Time Taken
• a = Acceleration

According to the question by using the formula we get,

$\implies \sf s =\: (65)(12) + \dfrac{1}{2} \times (2.75)(12)^2$

$\implies \sf s =\: 65 \times 12 + \dfrac{1}{2} \times 2.75 \times (12 \times 12)^2$

$\implies \sf s =\: 780 + \dfrac{1}{\cancel{2}} \times {\cancel{396}}$

$\implies \sf s =\: 780 + 198$

$\implies \sf\bold{\red{s =\: 978\: m}}$

${\small{\bold{\underline{\therefore\: The\: distance\: covered\: by\: body\: is\: 978\: m\: .}}}}$

✪ Solution No 2 :- ✪

Given :

• A body starts from rest and acquires a velocity of 18 km/h in 10 seconds.

To Find :-

• What is the acceleration of the body.

Solution :-

First, we have to convert the final velocity km/h into m/s :

$\leadsto \sf Final\: Velocity =\: 18\: km/h$

$\leadsto \sf Final\: Velocity =\: {\cancel{18}} \times \dfrac{5}{\cancel{18}}\: m/s$

$\leadsto \sf\bold{\purple{Final\: Velocity =\: 5\: m/s}}$

Now, we have to find the acceleration of the body :

As we know that :

$\clubsuit$ First Equation Of Motion Formula :

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

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

Given :

• Final Velocity = 5 m/s
• Initial Velocity = 0 m/s
• Time Taken = 10 seconds

According to the question by using the formula we get,

$\implies \sf 5 =\: 0 + a(10)$

$\implies \sf 5 - 0 =\: 10a$

$\implies \sf 5 =\: 10a$

$\implies \sf \dfrac{5}{10} =\: a$

$\implies \sf 0.5 =\: a$

$\implies \sf\bold{\red{a =\: 0.5\: m/s^2}}$

${\small{\bold{\underline{\therefore\: The\: acceleration\: of\: the\: body\: is\: 0.5\: m/s^2\: .}}}}$

✪ Solution No 3 :- ✪

Given :

• A body of mass 10 kg is moving with a velocity of 5 m/s attains of 35 m/s in 5 seconds.

To Find :-

• What is the force applied on body to change its velocity.

Solution :-

First, we have to find the acceleration :

As we know that :

$\clubsuit$ First Equation Of Motion Formula :

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

Given :

• Final Velocity = 35 m/s
• Initial Velocity = 5 m/s
• Time Taken = 5 seconds

According to the question by using the formula we get,

$\leadsto \sf 35 =\: 5 + a(5)$

$\leadsto \sf 35 - 5 =\: 5a$

$\leadsto \sf 30 =\: 5a$

$\leadsto \sf \dfrac{30}{5} =\: a$

$\leadsto \sf 6 =\: a$

$\leadsto \sf\bold{\purple{a =\: 6\: m/s^2}}$

Now, we have to find the force applied on body to change its velocity :

As we know that :

$\clubsuit$ Force Formula :

$\mapsto \sf\boxed{\bold{\pink{Force =\: Mass \times Acceleration}}}$

Given :

• Mass = 10 kg
• Acceleration = 6 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf Force =\: 10 \times 6$

$\longrightarrow \sf\bold{\red{Force =\: 60\: N}}$

${\small{\bold{\underline{\therefore\: The\: force\: applied\: on\: body\: to\: change\: its\: velocity\: is\: 60\: N\: .}}}}$

✪ Solution No 4 :- ✪

Given :

• A bus decreases its speed from 80 km/h to 60 km/h in 5 seconds.

To Find :-

• What is the acceleration of the bus.

Solution :-

First, we have to change the initial velocity and final velocity km/h into m/s :

❒ In case of initial velocity :

$\leadsto \sf Initial Velocity = 80\: km/h$

$\leadsto \sf Initial\: Velocity =\: 80 \times \dfrac{5}{18}\: m/s$

$\leadsto \sf Initial\: Velocity =\: \dfrac{400}{18}\: m/s$

$\leadsto \sf\bold{\purple{Initial Velocity =\: 22.22\: m/s}}$

❒ In case of final velocity :

$\leadsto \sf Final\: Velocity =\: 60\: km/h$

$\leadsto \sf Final\: Velocity =\: 60 \times \dfrac{5}{18}\: m/s$

$\leadsto \sf Final\: Velocity =\: \dfrac{300}{18}\: m/s$

$\leadsto \sf\bold{\purple{Final\: Velocity =\: 16.67\: m/s}}$

Now, we have to find the acceleration of the bus :

As we know that :

$\clubsuit$ First Equation Of Motion Formula :

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

Given :

• Final Velocity = 16.67 m/s
• Initial Velocity = 22.22 m/s
• Time Taken = 5 seconds

According to the question by using the formula we get,

$\implies \sf 16.67 =\: 22.22 + a(5)$

$\implies \sf 16.67 - 22.22 =\: 5a$

$\implies \sf - 5.55 =\: 5a$

$\implies \sf \dfrac{- 5.55}{5} =\: a$

$\implies \sf - 1.11 =\: a$

$\implies \sf\bold{\red{a =\: - 1.11\: m/s^2}}$

${\small{\bold{\underline{\therefore\: The\: acceleration\: of\: the\: bus\: is\: - 1.11\: m/s^2\: .}}}}$

Answer 2

Questions

1) The velocity of a body changes from 65 m/s - 98 m/s by time of 12s. What is the distance covered by the body if acceleration is 2.75 m/s²?

2) A body starts from rest and acquires a velocity of 18 km/h in 10 s. The acceleration of the body is?

3) a body of mass 10 kg is moving with a velocity of 5 m/s attains of 35m/s in 5 seconds, find the force applied on the body to change its velocity?

4) A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

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Answers

1. Intial Velocity (u) = 65 m/s

Final Velocity (v) = 98 m/s

Time (t) = 12 seconds

Acceleration (a) = 2.75 m/s²

Distance covered (s) = ?

To find the distance covered, we'll use 2nd Equation of Motion.

2nd Equation of Motion → $s = ut+\dfrac{1}{2}at^{2}$

Substitute the required values and find the distance covered.

⇒ $s = 65\times 12 + \dfrac{1}{2}(2.75)(12)^{2}$

⇒ $s = 780+\dfrac{1}{2}(2.75)(144)$

⇒ $s = 780+72(2.75)$

⇒ $s = 780+198$

⇒ $s = 978\ m$

∴ The distance covered by the body is 987 m.

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2. Initial Velocity (u) = 0 m/s

Final Velocity (v) = 18 km/h = 5 m/s

Time (t) = 10 seconds

Acceleration (a) = ?

We'll apply the 1st Equation of motion to find the acceleration.

1st Equation of Motion → $v=u+at$

Substitute the required values and find the acceleration.

⇒ 5 = 0 + a(10)

⇒ 5 = 10a

⇒ a = 5/10

⇒ a = 0.5 m/s²

∴ The acceleration of the body is 0.5 m/s²

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3. Mass of body (m) = 10 kg

Initial Velocity (u) = 5 m/s

Final Velocity (v) = 35 m/s

Time = 5 seconds

Force applied (F) = ?

Force = ma {here 'm' is mass and 'a' is acceleration}

To find the force applied we must find the acceleration of the body first.

To find the acceleration of the body, we'll use the 1st Equation of Motion.

1st Equation of Motion → $v=u+at$

Substitute the required values and find the acceleration.

⇒ 35 = 5 + a(5)

⇒ 35 = 5 + 5a

⇒ 5a = 35 - 5

⇒ 5a = 30

⇒ a = 30/5

⇒ a = 6 m/s²

∴ The acceleration of the body is 6 m/s²

Force = Mass × Acceleration

Force = 10 × 6

Force = 60 N

∴ The force applied on the body to change its velocity is 60N.

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4. Initial Velocity (u) = 80 km/h = 22.22 m/s

Final Velocity (v) = 60 km/h = 16.66 m/s

Time (t) = 5 seconds

Acceleration (a) = ?

We'll apply the 1st Equation of motion to find the acceleration.

1st Equation of Motion → $v=u+at$

Substitute the required values and find the acceleration.

⇒ 16.66 = 22.22 + a(5)

⇒ 22.22 + 5a = 16.66

⇒ 5a = 16.66 - 22.22

⇒ 5a = -5.56

⇒ a = -5.56/5

⇒ a = -1.112 m/s²

∴ The acceleration of the bus is -1.112 m/s²

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