Question

A car initially at rest start moving with acceleration 0.5m/s2cocovers a distance of 25 metre calculate the time required to cover this distance and final velocity of car

Answer 1

Answer:

From equation of motion v2=u 2 +2a

v 2=(0) 2+2×0.5×25or

v 2 =25

or

Final velocity

v= 25=5ms −1

Explanation:

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Answer 2

$\:\:\green{\underline{\underline{\textsf{\textbf{\purple{\:\:\:Hola }}}}}}$⚘⚘

${\underline{\underline{\mathfrak{\red{\:\:\:Given\:\:\:}}}}}$

$\tiny\:\:\:\bullet\:\:\sf{Initial\: velocity\:(u) = 0\:m/s}$

$\tiny\:\:\:\bullet\:\:\sf{Acceleration \:(a) = 0.5\:m/s^{2}}$

$\tiny\:\:\:\bullet\:\:\sf{Distance\: travelled\:(s) = 25\:m}$

${\underline{\underline{\mathfrak{\orange{\:\:\:To\:Find\:\:\:}}}}}$

$\tiny\:\:\:\bullet\:\:\sf{Time\:taken\:(t)}$

$\tiny\:\:\:\bullet\:\:\sf{Final\: velocity\:(v)}$

${\underline{\underline{\mathfrak{\blue{\:\:\: Solution\:\:\:}}}}}$

$\tiny\bigstar\:{\underline{\sf{By\: using\:the\:2^{nd}\: equation\: of\: motion}}}$

$\:\:\:\leadsto\:\:\sf{s = ut + \frac{1}{2} at^{2}}$

$\:\:\:\leadsto\:\:\sf{25 = 0 \times (t) + \frac{1}{2} \times 0.5\times (t)^{2}}$

$\:\:\:\leadsto\:\:\sf{25 = 0.25 \:t^{2}}$

$\:\:\:\leadsto\:\:\sf{t^{2}=\frac{25}{0.25}}$

$\:\:\:\leadsto\:\:\sf{t^{2}=100}$

$\:\:\:\leadsto\:\:\sf\pink{t= 10\:s}$

$\tiny\bigstar\:{\underline{\sf{By\: using\:the\:1^{st}\: equation\: of\: motion}}}$

$\:\:\:\leadsto\:\:\sf{v = u+at}$

$\:\:\:\leadsto\:\:\sf{v = 0+0.5\times 10}$

$\:\:\:\leadsto\:\:\sf\pink{v = 5\:m/s}$

$\bigstar\:{\underline{\sf{\purple{\:\:Hence\:\:}}}}$

$\tiny\:\:\:\bullet\:\:\sf{Time\:taken\:(t)= 10\:s}$

$\tiny\:\:\:\bullet\:\:\sf{Final\: velocity\:(v) = 5\:m/s}$