Question

A body starts form rest and move with uniform acceleration 5m/s for 8 seconds from that time the acceleration ceases the distant ve covered in 12 seconds starting from rest is

Answer 1

$\large{\underline{\underline{\mathfrak{Answer :}}}}$

• Distance traveled in 12s is 360m

$\rule{200}{0.5}$

$\underline{\underline{\mathfrak{Step-By-Step-Explanation :}}}$

Given :

• Acceleration (a) = 5 m/s² for 8s
• Initial velocity (u) = 0 m/s

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To Find :

• Distance covered in 12 sec from rest

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Solution :

• Distance traveled in 8s

Use 2nd equation of motion :

$\large{\boxed{\sf{s \: = \: ut \: + \: \dfrac{1}{2} at^2}}} \\ \\ \implies {\sf{s \: = \: 0 \: \times \: 8 \: + \: \dfrac{1}{2} \: \times \: 5 \: \times \: 8^2}} \\ \\ \implies {\sf{s \: = \: 0 \: + \: \dfrac{5 \: \times \: 64}{2}}} \\ \\ \implies {\sf{s \: = \: 5 \: \times \: 32}} \\ \\ \implies {\sf{s \: = \: 160 \: m}} \\ \\ \underline{\sf{\therefore \: Distance \: traveled \: in \: first \: 8s \: is \: 160 \: m}}$

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Now, we have to find velocity after 8s (means final velocity)

Use 1st equation of motion :

$\large{\boxed{\sf{v \: = \: u \: + \: at}}} \\ \\ \implies {\sf{v \: = \: 0 \: + \: 5 \: \times \: 8}} \\ \\ \implies {\sf{v \: = \: 40}} \\ \\ \underline{\sf{\therefore \: Velocity \: after \: 8s \: is \: 40 \: ms^{-1}}}$

• Now, use Formula for velocity for calculating distance in last 4s

$\large{\boxed{\sf{v \: = \: \dfrac{distance}{time}}}} \\ \\ \implies {\sf{distance \: = \: 4 \: \times \: 40}} \\ \\ \implies {\sf{Distance \: = \: 160 \: m}} \\ \\ \underline{\sf{\therefore \: Distance \: traveled \: in \: last \: 4s \: is \: 160 \: m}}$

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• Now, we have to find total distance in 12s

⇒Total Distance = Distance in 8s + Distance in 4s

⇒Total Distance = 160 + 160

⇒Total Distance = 320 m

$\therefore$ Total Distance traveled in 12s is 320 m