Question

For an object moving with an initial speed of 72 km/hr, what is the time required to stop it with a deceleration of 2m/s^2​

Answer 1

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

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

⋆ There is an object. The object is moving with an initial speed of 72 km/hr. Now this question says that we have to find out the time that is required stop the object with a deceleration or retardation (that is inverse of acceleration) of 2 m/s sq.

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

$\sf According \: to \: question \begin{cases} & \sf{Initial \: velocity \: = \bf{72 \: km/h}} \\ \\ & \sf{Deceleration \: = \bf{2 \: m/s^{2}}} \\ \\ & \sf{Final \: velocity \: = \bf{0 \: m/s}} \\ \\ & \sf{Time \: = \bf{?}} \end{cases}$

Don't be confused! Final velocity came as zero because we have stop the object means the object is at rest.

${\bigstar \:{\pmb{\sf{\underline{Knowledge \: requied...}}}}}$

⋆ The SI unit of acceleration and deceleration is m/s sq.

⋆ The velocities are shown by the unit m/s always.

• In question initial velocity or initial speed is in km/h so we have to change it into m/s as we take final velocity as 0 m/s!

⋆ Declaration is inverse of acceleration so it be in negative always.

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

⋆ First equation of motion that is given by the mentioned idea:

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

(Where, v denotes final velocity , u denotes initial velocity , a denotes acceleration and t denotes time)

⋆ Formula to change kilometres into metres is mentioned below:

${\small{\underline{\boxed{\sf{1 \: kilometre \: = 1000 \: metres}}}}}$

⋆ Formula to change hours into seconds is mentioned below

⠀⠀⠀⠀⠀⠀${\small{\underline{\boxed{\sf{1 \: hr \: = 3600 \: seconds}}}}}$

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

~ Firstly changing kilometres into metres and hours into seconds that is let us change km/h into m/s!

$:\implies \sf 72 \dfrac{km}{hr} \times \dfrac{1000 \: km}{1 \: m} \times \dfrac{1 \: hr}{3600 \: s} \\ \\ :\implies \sf 72 \dfrac{km}{hr} \times \dfrac{10\cancel{00} \: km}{1 \: m} \times \dfrac{1 \: hr}{36\cancel{00} \: s} \qquad (Cancelling \: terms) \\ \\ :\implies \sf 72 \dfrac{km}{hr} \times \dfrac{10 \: km}{1 \: m} \times \dfrac{1 \: hr}{36 \: s}\\ \\ :\implies \sf 72 \dfrac{km}{hr} \times \dfrac{{\cancel{10 \: km}}}{1 \: m} \times \dfrac{1 \: hr}{\cancel{{36 \: s}}}\qquad (Cancelling \: terms)\\ \\ :\implies \sf 72 \dfrac{km}{hr} \times \dfrac{5 \: km}{1 \: m} \times \dfrac{1 \: hr}{18 \: s} \\ \\ :\implies \sf \cancel{72} \dfrac{km}{hr} \times \dfrac{5\: km}{1 \: m} \times \dfrac{1 \: hr}{\cancel{{18\: s}}}\qquad (Cancelling \: terms)\\ \\ :\implies \sf 36 \dfrac{km}{hr} \times \dfrac{5 \: km}{1 \: m} \times \dfrac{1 \: hr}{9 \: s}\\ \\ :\implies \sf \cancel{36} \dfrac{km}{hr} \times \dfrac{5 \: km}{1 \: m} \times \dfrac{1 \: hr}{\cancel{{9 \: s}}} \qquad (Cancelling \: terms)\\ \\ :\implies \sf 4 \dfrac{km}{h} \times \dfrac{5 \: km}{1 \: m} \times \dfrac{1 \: hr}{1 \: s}\\ \\ :\implies \sf 20 \: m/s \\ \\ {\pmb{\tt{Henceforth, \: converted}}}$

~ Now let's use formula of first law of equation to get our answer!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 20 + (-2)(t) \\ \\ :\implies \sf 0 = 20 + (-2t) \\ \\ :\implies \sf 0 = 20 - 2t \\ \\ :\implies \sf 0 - 20 = -2t \\ \\ :\implies \sf -20 = -2t \\ \\ :\implies \sf \cancel{-}20 = \cancel{-}2t \qquad (Cancelling \: signs) \\ \\ :\implies \sf 20 = 2t \\ \\ :\implies \sf \dfrac{20}{2} = t \\ \\ :\implies \sf \cancel{\dfrac{20}{2}} = t \qquad (Cancelling \: terms) \\ \\ :\implies \sf 10 = t \\ \\ :\implies \sf Time \: = 10 \: seconds$

Henceforth, 10 seconds are needed to stop the object...