Question

A body travels 2 m in the first two second and 2.20 m in the
next 4 second with uniform deceleration. The velocity of the
body at the end of 9 second is
(a) – 10 ms⁻¹ (b) – 0.20 ms⁻¹
(c) – 0.40 ms⁻¹ (d) – 0.80 ms⁻¹

Answer 1

Answer:

$A to B</p><p>2=u×2+21×a×2×2⇒1=u+a</p><p>A to C</p><p>4.20=u×6+21×a×6×6⇒0.7=u+3a</p><p></p><p></p><p>Let the Body travel from A to B in 2 s for a distance of 2 m</p><p>Let the Body travel from B to C for next 4s for a distance of 2.20 m</p><p>Velocity after 9 s=?</p><p>For AtoB</p><p>WKT, S=ut+21at2⇒2=2u+21a∗2∗2⇒1=u+a...(1)</p><p>For B to C</p><p>WKT, S=ut+21at2⇒4.20=6u+21a∗6∗6⇒0.7=u+3a...(2)</p><p>From (1) and (2), we get,</p><p>2a=−0.3⇒a=−0.15,−ve as it is decreasing acceleration .</p><p>u=1−a⇒u=1+0.15=1.15</p><p>Now, velocity at t=9s=v=u+at⇒v=1.15−0.15∗9=1.15−1.35=−0.2m/s</p><p>velocity is negative as it is decreasing. \:$

$A to B</p><p>2=u×2+21×a×2×2⇒1=u+a</p><p>A to C</p><p>4.20=u×6+21×a×6×6⇒0.7=u+3a</p><p></p><p></p><p>Let the Body travel from A to B in 2 s for a distance of 2 m</p><p>Let the Body travel from B to C for next 4s for a distance of 2.20 m</p><p>Velocity after 9 s=?</p><p>For AtoB</p><p>WKT, S=ut+21at2⇒2=2u+21a∗2∗2⇒1=u+a...(1)</p><p>For B to C</p><p>WKT, S=ut+21at2⇒4.20=6u+21a∗6∗6⇒0.7=u+3a...(2)</p><p>From (1) and (2), we get,</p><p>2a=−0.3⇒a=−0.15,−ve as it is decreasing acceleration .</p><p>u=1−a⇒u=1+0.15=1.15</p><p>Now, velocity at t=9s=v=u+at⇒v=1.15−0.15∗9=1.15−1.35=−0.2m/s</p><p>velocity is negative as it is decreasing. \:$

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Answer 2

Explanation:

the answer is 1/25 m/sec by using s= ut + 1/2 at^2

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