Question

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Here is a question to check out your physics. Quality and correct answer needed. Please help :-

1.) A body starts with an initial velocity of 10 m/sec and acceleration 5 m/sec². Find distance covered by it in 5 seconds.

2.) An object moving along a straight line is initially moving at a velocity of +12 m/sec but is accelerating so that at a time 4 seconds later it is moving at the same speed, but in the opposite direction.
What is the object's average acceleration during the 4 seconds ?

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Answer 1

Answer :-

$\: \\ \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$

Here the concept of Equations of Motions and Acceleration - Time relationship has been used.

We know that if we know time(t), initial velocity (u), final velocity (v), displacement (s) and acceleration(a), we can apply their values and satisfy the equation.

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★ Formula Used :-

$\: \\ \large{\boxed{\boxed{\sf{s\:\: = \: \bf{ut \: + \: \dfrac{1}{2} \: \times \: at^{2}}}}}}$

This is the Second Equation of Motion.

$\: \\ \large{\boxed{\boxed{\sf{Acceleration,\:(a) \: = \: \bf{\dfrac{v_{(v_{f})}\: - \: u_{(v_{i})}}{t}}}}}}$

This is the Acceleration - Time Relationship.

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★ Solution :-

1.) A body starts with an initial velocity of 10 m/sec and acceleration 5 m/sec². Find distance covered by it in 5 seconds.

Answer.) Given,

» Initial velocity of the body = u = 10 m/sec

» Acceleration of the body = a = 5 m/sec²

» Time taken by the body = t = 5 seconds

Here we no need to find final velocity (v) in this case because w we can simply apply the values in the Second Equation of Motion and get our answer.

Then,

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: s\:\: = \: \bf{ut \: + \: \dfrac{1}{2} \: \times \: at^{2}}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: s\:\: = \: \bf{(10\: m\cancel{sec^{-1}} \: \times \: 5 \: \cancel{sec}) \: + \: \dfrac{1}{2} \: \times \: (5\: m\cancel{sec^{-2}})(5^{2} \: \cancel{sec^{2}})}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: s\:\: = \: \tt{10 \: \times \: 5 \: + \: \dfrac{1}{2} \: \times \: 5 \: \times \: 25 \: \: m \: \: = \: \bf{50 \; + \; \dfrac{1}{2} \: \times \: 125 \: \; \: m}}}}$

On taking the LCM, we get,

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: s\:\: = \: \bf{\dfrac{100\: + \: 125}{2} \: \;\:m}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: s\:\: = \: \tt{\dfrac{\cancel{225}}{\cancel{2}} \: \;\: m \: \:= \: \: \underline{\underline{\bf{112.5 \: \; m}}}}}}$

$\: \: \\ \large{\underline{\underline{\rm{Thus, \: the \: distance \: travelled \: by \: the \: body \: is \: \: \boxed{\bf{112.5 \: m}}}}}}$

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2.) An object moving along a straight line is initially moving at a velocity of +12 m/sec but is accelerating so that at a time 4 seconds later it is moving at the same speed, but in the opposite direction.

What is the object's average acceleration during the 4 seconds ?

Answer.) Given,

» Initial velocity of the object = u = + 12 m/sec

Here +ve sign denotes its direction.

We are given that the final velocity of the body has speed equal to the initial speed but in opposite direction. So, its direction will be opposite to +ve.

» Final Velocity of the object = v = - 12 m/sec

» Time taken to accelerate = t = 4 seconds

Then, by using Acceleration time relationship, we get,

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: Acceleration,\:(a) \: = \: \bf{\dfrac{v_{(v_{f})}\: - \: u_{(v_{i})}}{t}}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: Acceleration,\:(a) \: = \: \bf{\dfrac{- \: 12 \: msec^{-1} \: - \: (+ \: 12 \: msec^{-1})}{4 \: sec}}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: Acceleration,\:(a) \: = \: \bf{\dfrac{- \: 12 \: msec^{-1} \: - \: \: 12 \: msec^{-1}}{4 \: sec}}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: Acceleration,\:(a) \: = \: \tt{\dfrac{ \: - \: \: \cancel{24} \: msec^{-1}}{\cancel{4} \: sec} \: \: = \: \: \underline{\underline{\bf{- \: 6 \: msec^{-2}}}}}}}$

Here negative sign shows that the body is declarating. Thus this is retardation.

$\: \: \\ \large{\underline{\underline{\rm{Thus, \: the \: average \: acceleration \:of \: the \: object \: is \: \: \boxed{\bf{- \: 6 \: msec^{-2}}}}}}}$

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$\: \\ \qquad \: \large{\underbrace{\underbrace{\sf{More \: to \: know \: :-}}}}$

$\: \\ \sf{v \: - \: u \: \: = \: \: at}$

This is the first equation of motion.

$\: \\ \sf{v^{2} \: - \: u^{2} \: \: = \: \: 2as}$

This is the third equation of motion.

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Answer 2

Solution for 1 :

Given,

A body starts with an initial velocity of 10 m/sec and acceleration 5 m/sec².

To find,

Find distance covered by it in 5 seconds.

Explanation :

We have, u = 10 m/s ; a = 5 m/s² ; t = 5 s

Now using 2nd equation of motion :

⇒ s = ut + 1/2 at²

⇒ s = 10 * 5 + 1/2 * 5 * 5²

⇒ s = 50 + 1/2 * 5 * 25

⇒ s = 50 + 125/2

⇒ s = 50 + 62.5

⇒ s = 112.5 m

∴ Distance covered by it in 5 s = 112.5 m

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Solution for 2 :

Given,

An object moving along a straight line is initially moving at a velocity of +12 m/sec but is accelerating so that at a time 4 seconds later it is moving at the same speed, but in the opposite direction.

To find,

What is the object's average acceleration during the 4 seconds.

Explanation :

Initial velocity of object , u = +12 m/s

Final velocity of object , v = -12m/s

Time, t = 4s

Now using 1st equation of motion :

⇒ v = u + at

⇒ -12 = 12 + a * 4

⇒ -12 -12 = 4a

⇒ -24 = 4a

⇒ a = -24/4

⇒ a = -6 m/s²

∴ Average acceleration of object = -6 m/s²