Question

an object is moving with constant acceleration.initially it is travelling at 16m/s .three seconds later it is travelling at 10m/s .how far it travel during this time​

Answer 1

Given:-

• Initial velocity ,u = 16m/s

• Final velocity ,v = 10m/s

• Time taken ,t = 2 s

To Find:-

• Distance travelled by object,s

Solution:-

Firstly we calculate the Acceleration of the object.

Using 1st Equation of motion

• v = u +at

Substitute the value we get

→ 10 = 16 + a ×2

→ 10-16 = a×2

→ -6 = a×2

→ a = -6/2

→ a = -3m/s²

here, negative sign show retardation

Therefore the acceleration of the object is 3m/s²

Now, Using 3rd Equation of motion

• v² = u² +2as

Substitute the value we get

→ 10² = 16² + 2×(-3) ×s

→ 100 = 196 + (-6 ) ×s

→ 100 -196 = -6×s

→ -96 = -6×s

→ s = -96/-6

→ s = 96/6

→ s = 16m

Therefore, the distance covered by the car is 16 metres.

Answer 2

❏ GIVEN :

• Initial velocity (u) = 16 m/s

• Final velocity (v) = 10 m/s

• Time (t) = 2 sec

❏ TO FIND :

• Distance travelled by the object (s)

❏ SOLUTION :

✦ First we have to find the acceleration of the object.

✦ We Use:-

$➥ \large \underline{\boldsymbol \color{green}{First ~ equation ~ of ~ motion: -}}$

$\bigstar \: \large\color{blue} { \boxed{ \boxed{ \bf\color{blue}{v =u + at }}}}$

✯ WHERE :

• v = final velocity

• u = initial velocity

• a = acceleration

• t = time

━━━━━━━━━━━━━━━━━━━━━━━━━━

♣ Putting all values :

➙ 10 = 16 + a × 2

➙ 10 = 16 + 2a

➙ 10 - 16 = 2a

➙ -6 = 2a

➙ 2a = -6

➙ a = -(6/2)

➜ a = -3

━━━━━━━━━━━━━━━━━━━━━━━━━━

⛬ Acceleration = 3 m/s² ( negative sign show retardation)

✦ Now to find the distance travelled by the object :-

✦ We Use :-

$➥ \large \underline{\boldsymbol \color{green}{Third~ equation ~ of ~ motion: -}}$

$\bigstar \: \large\color{blue} { \boxed{ \boxed{ \bf\color{blue}{v^{2} =u^{2} + 2as }}}}$

✯ WHERE :

v = final velocity

u = initial velocity

a = acceleration

s = distance covered

━━━━━━━━━━━━━━━━━━━━━━━━━━

♣ Putting all values :

➙ 10² = 16² + 2 × (-3) × s

➙ 100 = 196 + (-6) × s

➙ 100 - 196 = (-6) × s

➙ - 96 = -6s

➙ -6s = -96

$➜\: \bf{s = \dfrac{ \cancel{-} 96}{ \cancel{-} 6} = \cancel\dfrac{ 96}{6} = 16 }$

━━━━━━━━━━━━━━━━━━━━━━━━━━

⛬ the distance covered by the car = 16 m

❏ ANSWER :

The distance covered by the car is 16 m.

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✪ More information ✪

✦ First equation of motion :-

$\large\color{grey} { \boxed {\underline{ \bf\color{grey}{v =u + at }}}}$

Where :-

• v = final velocity

• u = initial velocity

• a = acceleration

• t = time taken

✦ Second equation of motion :-

$\large\color{grey} { \boxed {\underline{ \bf\color{grey}{ s\: =\: ut\: +\: ½\: at² }}}}$

Where :-

• s = distance covered

• u = initial velocity

• t = time taken

✦ Third equation of motion :-

$\large\color{grey} { \boxed {\underline{ \bf\color{grey}{ v²\: -\: u²\: =\: 2as}}}}$

Where :-

• v = final velocity

• u = initial velocity

• a = acceleration

• s = distance covered

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