Question

${\bf{\orange{Subject }{\leadsto{\sf{Motion \: and \: Force}}}}}$
${\bf{\orange{Subject} {\leadsto{\sf{Physics}}}}}$

${\large{\red{\mathbb{\mathfrak{\underline{Question :-}}}}}}$
An object starting from rest travels a distance of 20m in first 2s and 160m in 4s.What will be the velocity after 7s of start.

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Answer 1

Answer:

$\bold{The \;velocity \;of \;given\; object\; after\; 7s \;from\; the\; starting\; point\; is\; 70 \;m/s .}$

Explanation:

Given :

$\Rightarrow \bold{Distance_1=20m}$

$\Rightarrow \bold{Time\;taken_1=2s}$

$\Rightarrow \bold{Distance_1=160m}$

$\Rightarrow \bold{Time_2=7s}$

To find :

$\Rightarrow \bold{Velocity\;after\;7\;seconds\;of\;start.}$

Solution :

Using first equation of motion:

$\boxed{\bold{v=u+at}}$

Using second equation of motion:

$\boxed{\bold{s=ut+\frac{1}{2}at^2 }}$

Finding the acceleration₁:

$\rightarrow \bold{20=0(2)+\frac{1}{2}a_1(2)^2 }$

$\rightarrow \bold{20=\frac{1}{2}a_14}$

$\rightarrow \bold{20=2a_1}$

$\rightarrow \bold{a_1=10m/s^2}$

Finding the final velocity:

$\rightarrow \bold{v=0+10(2)}$

$\rightarrow \bold{v=20m/s}$

Hence, the final velocity (20m/s) on given body for the next 4s:

Find the acceleration₂:

$\rightarrow \bold{160=20(4)+\frac{1}{2}a_2(4)^2 }$

$\rightarrow \bold{160=80+8a_2}$

$\rightarrow \bold{80=8a_2}$

$\rightarrow \bold{a_2=10m/s^2}$

Find the velocity for whole journey:

$\rightarrow \bold{v=0+10(7)}$

$\rightarrow \bold{v=70m/s}$

∴ $\bold{The \;velocity \;of \;given\; object\; after\; 7s \;from\; the\; starting\; point\; is\; 70 \;m/s .}$

Answer 2

Answer:

$Thevelocityofgivenobjectafter7sfromthestartingpointis70m/s.</p><p>Explanation:</p><p>Given :</p><p>\Rightarrow \bold{Distance_1=20m}⇒Distance1=20m</p><p>\Rightarrow \bold{Time\;taken_1=2s}⇒Timetaken1=2s</p><p>\Rightarrow \bold{Distance_1=160m}⇒Distance1=160m</p><p>\Rightarrow \bold{Time_2=7s}⇒Time2=7s</p><p>To find :</p><p>\Rightarrow \bold{Velocity\;after\;7\;seconds\;of\;start.}⇒Velocityafter7secondsofstart.</p><p>Solution :</p><p>Using first equation of motion:</p><p>\boxed{\bold{v=u+at}}v=u+at</p><p>Using second equation of motion:</p><p>\boxed{\bold{s=ut+\frac{1}{2}at^2 }}s=ut+21at2</p><p>Finding the acceleration₁:</p><p>\rightarrow \bold{20=0(2)+\frac{1}{2}a_1(2)^2 }→20=0(2)+21a1(2)2</p><p>\rightarrow \bold{20=\frac{1}{2}a_14}→20=21a14</p><p>\rightarrow \bold{20=2a_1}→20=2a1</p><p>\rightarrow \bold{a_1=10m/s^2}→a1=10m/s2</p><p>Finding the final velocity:</p><p>\rightarrow \bold{v=0+10(2)}→v=0+10(2)</p><p>\rightarrow \bold{v=20m/s}→v=20m/s</p><p>Hence, the final velocity (20m/s) on given body for the next 4s:</p><p>Find the acceleration₂:</p><p>\rightarrow \bold{160=20(4)+\frac{1}{2}a_2(4)^2 }→160=20(4)+21a2(4)2</p><p>\rightarrow \bold{160=80+8a_2}→160=80+8a2</p><p>\rightarrow \bold{80=8a_2}→80=8a2</p><p>\rightarrow \bold{a_2=10m/s^2}→a2=10m/s2</p><p>Find the velocity for whole journey:</p><p>\rightarrow \bold{v=0+10(7)}→v=0+10(7)</p><p>\rightarrow \bold{v=70m/s}→v=70m/s</p><p>∴ \bold{The \;velocity \;of \;given\; object\; after\; 7s \;from\; the\; starting\; point\; is\; 70 \;m/s .}Thevelocityofgivenobjectafter7sfromthestartingpointis70m/s.</p><p>$

Explanation:

This is the answer.