Question

an object starting from rest travel 20m in first 2s and 160 m in next 4s .find the velocity of object after 7 sec from the start​

Answer 1

${\large{\pmb{\sf{\underline{RequirEd \: Solution...}}}}}$

${\bigstar \:{\pmb{\sf{\underline{Understanding \: the \: question...}}}}}$

⋆ This question says that we have to find out the velocity of the given object after 7 seconds from the start when the object starts from the rest. Afterthat it travels 20 metres in first 2 second and 160 metres in next 4 seconds.

${\bigstar \:{\pmb{\sf{\underline{Provided \: that...}}}}}$

$\sf According \: to \: statement \begin{cases} & \sf{Initial \: velocity \: = \bf{0 \: m/s}} \\ \\ & \sf{Distance_1 \: = \bf{20 \: metres}} \\ \\ & \sf{Distance_2 \: = \bf{160 \: metres}} \\ \\ & \sf{Time_1 \: = \bf{2 \: seconds}} \\ \\ & \sf{Time_2 \: = \bf{4 \: seconds}} \end{cases}$

Don't be confused! Initial velocity came as zero because the object starts from rest.

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We are asked to find that the velocity of the given object after 7 seconds.

${\bigstar \:{\pmb{\sf{\underline{Using \: concepts...}}}}}$

⋆ Formula to find out the total distance.

⋆ Formula to find out the total time.

⋆ First equation of motion.

⋆ Second equation of motion.

${\bigstar \:{\pmb{\sf{\underline{Using \: formulas...}}}}}$

${\small{\underline{\boxed{\sf{\star \: Total \: distance \: = Distance_1 \: + Distance_2}}}}}$

${\small{\underline{\boxed{\sf{\star \: Total \: time \: = Time_1 \: + Time_2}}}}}$

⋆ First equation of motion is given by

${\small{\underline{\boxed{\sf{v \: = u \: + at}}}}}$

⋆ Second equation of motion is given by

${\small{\underline{\boxed{\sf{s \: = ut \: + \dfrac{1}{2} \times at^2}}}}}$

• (Where, v denotes final velocity , u denotes initial velocity , a denotes acceleration , t denotes time , s denotes displacement or distance or height)

${\bigstar \:{\pmb{\sf{\underline{Full \; Solution...}}}}}$

~ Firstly let us find total time

$:\implies \sf Total \: time \: = Time_1 \: + Time_2 \\ \\ :\implies \sf Total \: time \: = 2 + 4 \\ \\ {\pmb{\sf{:\implies Total \: time \: = 6 \: seconds}}}$

~ Now let us find total distance

$:\implies \sf Total \: distance \: = Distance_1 \: + Distance_2 \\ \\ :\implies \sf Total \: distance \: = 20 + 160 \\ \\ {\pmb{\sf{:\implies Total \: distance \: = 180 \: metres}}}$

~ Now let us use second equation of motion to find acceleration

$:\implies \sf s \: = ut \: + \dfrac{1}{2} \times at^2 \\ \\ :\implies \sf 180 = 0(6) + \dfrac{1}{2} \times a(6)^{2} \\ \\ :\implies \sf 180 = 0(6) + \dfrac{1}{2} \times a(36) \\ \\ :\implies \sf 180 = 0(6) + \dfrac{1}{2} \times 36a \\ \\ :\implies \sf 180 = 0 + \dfrac{1}{2} \times 36a \\ \\ :\implies \sf 180 = \dfrac{1}{2} \times 36a \\ \\ :\implies \sf 180 = \dfrac{1}{\cancel{{2}}} \times \cancel{36}a \: (Cancelling) \\ \\ :\implies \sf 180 = 1 \times 18a \\ \\ :\implies \sf 180 = 18a \\ \\ :\implies \sf \dfrac{180}{8} \: = a \\ \\ :\implies \sf \cancel{\dfrac{180}{8}} \: = a \: (Cancelling) \\ \\ :\implies \sf 10 \: = a \\ \\ :\implies \sf a \: = 10 \: ms^{-2} \\ \\ {\pmb{\sf{:\implies Acceleration \: = 10 \: ms^{-2}}}}$

~ Now let's use first equation of motion to find out the final velocity but at the place of time we have to put 7 seconds as it's provided to us - Read carefully!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf v \: = 0 + 10(7) \\ \\ :\implies \sf v \: = 0 + 70 \\ \\ :\implies \sf v \: = 70 \: m/s \\ \\ {\pmb{\sf{:\implies Final \: velocity \: = 70 \: m/s}}}$