Question

a scooter moving at the spped of 12 metre per second is stopped by applying breaks which produces uniform retardation of 0.6 metre per second square.how much distance will be covered by the scooter till it stops​

Answer 1

$\large \dag$ Question :-

A scooter moving at the spped of 12 metre per second is stopped by applying breaks which produces uniform retardation of 0.6 metre per second square.how much distance will be covered by the scooter till it stops.

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf  \green{Distance   \: Covered \:  by  \: Scooter \: is \: 120\: m }} }$

$\large \dag$ Step by step Explanation :-

Here we have :

• Initial Velocity = u = 12 m/s

• Acceleration = a = - 0.6 m/s²

( Retardation = Negative of Acceleration)

• Final Velocity = v = 0 m/s

⇝ Calculating Time Taken to stop :-)

❒ We Have 1st Equation of Motion as :

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{v = u + at}}}$

⏩ Applying 1st Equation of Motion ;

$:\longmapsto \rm 0 = 12 + ( - 0.6) \times t$

$:\longmapsto \rm 0 = 12 - 0.6t$

$:\longmapsto \rm 0.6t = 12$

$:\longmapsto \rm t = \frac{12}{0.6}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf t = 20 \: s} }}}$

⇝ Calculating Required Distance :-)

❒ We Have 2nd Equation of Motion as :

$\large \bf \red \bigstar \: \: \orange{ \underbrace{ \underline{s=ut+\dfrac{1}{2}at^2}}}$

⏩ Applying 2nd Equation of Motion ;

$:\longmapsto \rm s = 12 \times 20 + \frac{1}{2} \times ( - 0.6) \times {20}^{2}$

$:\longmapsto \rm s = 240 - 120$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf s = 120 \: m} }}}$

Therefore,

$\underline{\pink{\underline{\frak{\pmb{Distance \: Covered \: b \text y \: \text Scooter =120 \: m }}}}}$

Answer 2

$\red{\large\maltese \: \: \pmb{\mathbb{SOLUTION》}}}$

As the scooter is moving at the speed of 12 m/s so its initial velocity will be u = 12 m/s

Also the scooter produces a uniform retardation of -0.6 m/s² so acceleration will be a = - 0.6 m/s²

After applying brakes the scooter finally comes to rest to final velocity will be v = 0 m/s

$\large \bigstar$ Now by third equation of motion

$\large\dag\pmb{ \: {v}^{2} - {u}^{2} = 2as}$

$\implies \: 0 {}^{2} - 12 {}^{2} = 2 \times ( - 0.6) \times s$

$\implies \: 0 - 144 = - 1.2s$

$\implies \: s = \frac{ - 144}{ - 1.2}$

$\Large\red\implies \green{\pmb{ \underline{ \overline{ \boxed{s =120 \: m }}}}}$

Hence distance covered by scooter before it stops is 120 meter.