Question

Aaooo sbb


A particle has velocity of 5.5m/s due east and co.stant acceleration of 1m/s2 due east.Find The distance cornered by it in 6th second of motion

Answer 1

$\huge{\underline{\underline{\red{Answer}}}}$

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$\mathbb{\boxed{\pink{GIVEN}}}$

• velocity
• $= 2m \div {s}$
• Acceleration=
• $2m \div {s}^{2}$

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$\bf{\underline{\underline{\red{SOLUTION}}}}$

Here acceleration is negative as it's opposite to direction of motion.

at t= 5s

$s = ut + \frac{1}{2} a {t}^{2} \\ \implies \: s = 5.5 \times 5 - \frac{1}{2} \times 1 \times {5}^{2} \\ \implies \: s = 15m$

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at t=6s

$s = ut + \frac{1}{2} a {t}^{2} \\ \implies \: s = 5.5 \times 6- \frac{1}{2} \times 1 \times {6}^{2} \\ \implies \: s = 15m$

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at t=5.5s

$s = ut + \frac{1}{2} a {t}^{2} \\ \implies \: s = 5.5 \times 5.5- \frac{1}{2} \times 1 \times {5.5}^{2} \\ \implies \: s = 15.125m$

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Distance covered in 6th second

=(-2)×(15.125-15)m

= (-2)×0.125m

=0.25 m

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$\large{\boxed{\bold{\red{Answer=0.25m}}}}$