Question

$submersible\ is \ lowered \ into \ the \ sea \ at\ a\ speed\ of\ 4\ m\ s^{-2} \ and \ comes \ to \ rest \ with \ uniform \\retardation \ at \ a \ distance \ of \ 10 \ m \ below \ the \ surface. \ Calculate \ (a) its \ retardation \ and \\(b) \ the \ time \ it \ takes \ to \ come \ to \ rest.\\\\\\plz \answer\ with\ full \ explanation$

Answer 1

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

${\bigstar \:{\pmb{\sf{\underline{Question...}}}}}$

$\sf Submersible \ is \ lowered \ into \ the \ sea \ at \ a\ speed \ of \ 4\ m/s \\ \sf and \ comes \ to \ rest \ with \ uniform \ retardation \ at \ a \ distance \\ \sf of \ 10 \ m \ below \ the \ surface. \\ \ \ \ \sf Calculate \\ \sf (a) \: Its \ retardation \\ \sf (b) \: The \ time \ it \ takes \ to \ come \ to \ rest.$

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

⋆ This question says that there is a submersible that is a submarine and it is lowered into the sea at a speed of 4 m/s and then it comes to a rest with uniform retardation that is inverse of acceleration at a distance of 10 metres below the surface. Afterwards this question says that we have to calculate it's retardation and also the time that it takes to come to the rest. Let's solve this question!

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

$\sf According \: to \: statement \begin{cases} & \sf{Initial \: velocity \: = \bf{4 \: m/s}} \\ \\ & \sf{Distance \: = \bf{10 \: m}} \\ \\ & \sf{Retardation \: = \bf{?}} \\ \\ & \sf{Time \: to \: came \: at \: rest \: = \bf{?}} \end{cases}$

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

⋆ Acceleration: The term acceleration is a measure of the change in the velocity of an object per unit time.

• Deceleration is also known as retardation.

• Retardation is the inverse of acceleration. It is in negative.

SI unit of acceleration and deceleration both is m/s²

⋆ First equation of motion is given by

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

⋆ Third equation of motion is given by

⠀⠀⠀⠀⠀${\small{\underline{\boxed{\sf{2as \: = v^2 - u^2}}}}}$

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

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

~ Firstly let us find retardation by using the third equation of motion.

$:\implies \sf 2as \: = v^2 - u^2 \\ \\ :\implies \sf 2(a)(10) \: = (0)^{2} - (4)^{2} \\ \\ :\implies \sf 2(10a) \: = 0 - 16 \\ \\ :\implies \sf 20a \: = -16 \\ \\ :\implies \sf a \: = \dfrac{-16}{20} \\ \\ :\implies \sf a \: = \cancel{\dfrac{-16}{20}} \qquad (Cancelling) \\ \\ :\implies \sf a \: = \dfrac{-8}{10} \\ \\ :\implies \sf a \: = -0.8 \\ \\ :\implies \sf Acceleration \: = -0.8 \\ \\ :\implies \sf Retardation \: = -0.8 \: m/s^{2}$

Henceforth, we get -0.8 meter per second sq. as the retardation.

~ Now let's find the time that the submersible takes to come to rest.

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 \: = 4 + (-0.8)(t) \\ \\ :\implies \sf 0 \: = 4 + (-0.8t) \\ \\ :\implies \sf 0 \: = 4 -0.8t \\ \\ :\implies \sf 0 - 4 = -0.8t \\ \\ :\implies \sf -4 = -0.8t \\ \\ :\implies \sf 4 = 0.8t \\ \\ :\implies \sf \dfrac{4}{0.8} = t \\ \\ :\implies \sf \cancel{\dfrac{4}{0.8}} = t \qquad (Cancelling) \\ \\ :\implies \sf \dfrac{1}{0.2} = t \\ \\ :\implies \sf \dfrac{1 \times 10}{2} = t \\ \\ :\implies \sf \dfrac{10}{2} = t \\ \\ :\implies \sf \cancel{\dfrac{10}{2}} = t \qquad (Cancelling) \\ \\ :\implies \sf 5 = t \\ \\ :\implies \sf t = 5 \\ \\ :\implies \sf Time \: = 5 \: seconds$

Henceforth, 5 seconds is the time take by the submersible to come to rest.

Answer 2

for solution and answer please refer to the given image