Question

A car was travelling at a speed of 10 m/s. After the brakes were applied, it start to decelerate at a rate of 2 m/s ² and finally came to rest. Find the distance traveled by the car before coming to rest.
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Answer 1

Initial speed of the car   u=10 m/s

Initial speed of the car   u=10 m/sFinal speed of the car   v=0 m/s

Initial speed of the car   u=10 m/sFinal speed of the car   v=0 m/sAcceleration (or  retardation) of the car   a=−2 m/s

Initial speed of the car   u=10 m/sFinal speed of the car   v=0 m/sAcceleration (or  retardation) of the car   a=−2 m/sUsing   v^2−u^2=2aS

Initial speed of the car   u=10 m/sFinal speed of the car   v=0 m/sAcceleration (or  retardation) of the car   a=−2 m/sUsing   v^2−u^2=2aS∴   0−10^2=2(−2)S

$\sf⟹ S= \dfrac{100}{2 \times 2}=25 m</p><p>$

Answer 2

$\star \; {\underline{\boxed{\pmb{\green{\frak{ \; Given \; :- }}}}}}$

• Initial Velocity = 10 m/s
• Acceleration = 2 m/s²

$\star \; {\underline{\boxed{\pmb{\pink{\frak{ \; To \; Find \; :- }}}}}}$

• Distance Covered = ?

$\star \; {\underline{\boxed{\pmb{\red{\frak{ \; SolutioN \; :- }}}}}}$

$\dag$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ {v}^{2} - {u}^{2} = 2as }}}}}$

Where :

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance Covered

$\\ \qquad{\rule{150pt}{2pt}}$

$\dag$ Calculating the Distance Covered :

${\dashrightarrow{\qquad{\sf{ {v}^{2} - {u}^{2} = 2as }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ {0}^{2} - {10}^{2} = 2 \times ( - 2 ) \times s }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 0 - 100 = - 4 \times s }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ \cancel{-} 100 = \cancel{-} 4 \times s }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 100 = 4 \times s }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ \dfrac{100}{4} = s }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ \cancel\dfrac{100}{4} = s }}}} \\ \\ \\ \ {\qquad \; \; {\therefore \; {\underline{\boxed{\pmb{\purple{\sf{ Distance \; Covered = 25 \; m }}}}}}}}$

$\\ \qquad{\rule{150pt}{2pt}}$

$\dag$ Therefore :

❛❛ Car travelled 25 m before coming to rest . ❜❜

$\\ {\underline{\rule{300pt}{9pt}}}$