Question

A car travelling at 9 m/s accelerates and attains speed of 27 m/s in 5 second . Calculate the acceleration and distance covered in 5 second.​

Answer 1

Given :

• Initial velocity, u = 9 m/s
• Final velocity, v = 27 m/s
• Time taken, t = 5 seconds

To find :

• Acceleration, a
• Distance covered, s

According to the question,

➟ v = u + at

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time taken

➟ Substituting the values,

➟ 27 = 9 + a × 5

➟ 27 - 9 = 5a

➟ 18 = 5a

➟ 18 ÷ 5 = a

➟ 3.6 = a

• So,the acceleration is 3.6 m/s².

Now,

➟ s = ut + ½ at²

Where,

• s = Distance
• u = Initial velocity
• t = Time taken
• a = Acceleration

➟ Substituting the values,

➟ s = 9 × 5 + ½ × 3.6 × 5 × 5

➟ s = 45 + 1.8 × 25

➟ s = 45 + 45

➟ s = 90

• So, the distance covered in 5 seconds is 90 meters.

____________________

Answer 2

$\; \; \; \; \; \; \; \;{\Large{\bf{\underbrace{Required \; Solution}}}}$

${\large{\bold{\rm{\underline{Given \; that}}}}}$

✠ Initial velocity = 9 m/s

✠ Final velocity = 27 m/s

✠ Time = 5 seconds

${\large{\bold{\rm{\underline{To \; find}}}}}$

✠ Acceleration

✠ Distance

${\large{\bold{\rm{\underline{Solution}}}}}$

✠ Acceleration = 3.6 m/s²

✠ Distance = 90 m

${\large{\bold{\rm{\underline{Using \; concept}}}}}$

✠ Law of motions

${\large{\bold{\rm{\underline{Using \; formula}}}}}$

✠ v = u + at

Where,

⚕️ v denotes final velocity

⚕️ u denotes initial velocity

⚕️ a denotes acceleration

⚕️ t denotes time

✠ s = ut + ½ at²

Where,

⚕️ s denotes displacement or distance

⚕️ u denotes final velocity

⚕️ t denotes time

⚕️ a denotes acceleration

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

~ Let us find acceleration first,

⇢ v = u + at

⇢ 27 = 9 + a(5)

⇢ 27 = 9 + a × 5

⇢ 27 = 9 + 5a

⇢ 27 - 9 = 5a

⇢ 18 = 5a

⇢ 18 / 5 = a

⇢ 3.6 = a

⇢ a = 3.6 m/s²

${\frak{Henceforth, \: 3.6 \: m/s^{2} \: is \: acceleration}}$

~ Let us find the distance now,

⇢ s = ut + ½ at²

⇢ s = 9(5) + ½ (3.6)(5)²

⇢ s = 9 × 5 + ½ × 3.6 × (5)²

⇢ s = 9 × 5 + ½ × 3.6 × 25

⇢ s = 45 + 1.8 × 25

⇢ s = 45 + 45

⇢ s = 90 m

${\frak{Henceforth, \: 90 \: m \: is \: distance}}$

${\large{\bold{\rm{\underline{Additional \; information}}}}}$

☀ Newton's law of motion - Newton if the father of physics and he established the laws of motion in his book named ❛Principia❜ in the year 1667.

☀ Newton's first law of motion - Every body maintain it's maintain it's initial state of rest or motion with uniform speed on straight line unless an external force act on it.

◆ It's also known to be law of Galileo or law of inertia.

♛ Law of inertia = It is the property of a body by virtue of which the body opposes change in it's initial state of rest or motion with uniform speed on a straight line. There are 2 types of inertia –

⑴ Inertia of rest

⑵ Inertia of motion

Some examples of inertia –

⑴ When a coat /blanket is beaten by a stick the dust particles are removed.

⑵ When a car start suddenly the passenger bends backwards.

◆ Important - First law gives the definition of force.

♛ Force - It is that external cause which when acts on a body changes or tries to change the initial state of the body.

♛ Momentum - It is that property of a moving body and is defined of mass and velocity of the body. It's a vector quantity. SI unit is kgm/s.

$\: \: \: \: \: \: \:{\bf{Momentum = \: Mass \times Velocity}} \green\star$

☀ Newton's 2nd law of motion - The rate of change in momentum of a body is directly proportional to the applied force on the body and take place in the direction of force.

If f = force applied ; a = acceleration produced and m = mass of body the nF = ma.

◆ Newton's second law gives the magnitude of force.

◆ Newton's first law is contained in the second law.

☀ Newton's 3rd law of motion - To every action there is an equal and opposite reaction.

Some examples -

⑴ Motion of the rocket.

⑵ Recoil of the gun.