Question

HELP PLEASE

A bullet accelerates at a rate of 90,000m/s^2 from rest. Calculate its final velocity if it travels a distance of 0.5m while accelerating.

Answer 1

The final velocity of it travels a distance of 0.5 m while accelerating is v = 300 m/s.

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Step-by-step explanation :

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GivEn :

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• Initial velocity (u) = 0 m/s

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• Acceleration (a) = 90000 m/s²

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• Distance (s) = 0.5 m

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To find :

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• Final velocity (v) = ?

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SoluTion :

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⠀⠀⠀⠀Let's solve the question by using the equations of motion. It is given by,

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`\star \: { \boxed{ \sf{ \purple{{v}^(2) = {u}^(2) + 2as}}}}`

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`\: \: \: \: \: \: \: \: \: \: \: { \underline{ \sf{Substituting \: the \: values, }}}`

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`: \implies \sf \: \: \: {{v}^(2) = {u}^(2) + 2as} \\ \\ \\`

`: \implies \sf \: \: \: {v}^(2) = ( {0)}^(2) + 2(0.5)(90 000) \\ \\ \\`

`: \implies \sf \: \: \: {v}^(2) = 90 000 \\ \\ \\`

`: \implies \sf \: \: \:{ \boxed{ \bf{ \red{ v= 300 m/s }}}} \: \bigstar \\ \\`

Therefore, The final velocity of it travels a distance of 0.5 m while accelerating is v = 300 m/s .

~TheInsaneGirl