Question

On applying brakes a vehicles takes 10 seconds to stop, If it was running at a speed of 40m/s then how long distance will covered it before it stop?​

Answer 1

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Initial \:velocity\: (u) = 40 \:m/s}$

$\:\:\:\:\bullet\:\:\:\sf{Final \:velocity \:(v) = 0 \:m/s}$

$\:\:\:\:\bullet\:\:\:\sf{Time \:(t) = 10 \:seconds}$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Distance \ covered \ by \ vehicle \ before \ it \ stop}$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}}$

☯ Firstly, We'll find acceleration

$\dashrightarrow\:\: \sf{v = u + at}$

$\dashrightarrow\:\: \sf{0 = 40 + a \times 10 }$

$\dashrightarrow\:\: \sf{- 40 = 10a }$

$\dashrightarrow\:\: \sf{ \dfrac{ \cancel{- 40}}{\cancel{10}} = a}$

$\dashrightarrow\:\: \sf{a = - 4 \: m /{s}^{2} }$

☯ Now, Calculate the distance covered

$\dashrightarrow\:\: \sf{{v}^{2} = {u}^{2} + 2as }$

$\dashrightarrow\:\: \sf{{0}^{2} = {40}^{2} + 2 \times ( - 4) \times s }$

$\dashrightarrow\:\: \sf{ - 1600 = - 8s}$

$\dashrightarrow\:\: \sf{\dfrac{ \cancel{ - 1600}}{ \cancel{- 8}} = s }$

$\dashrightarrow\:\: {\boxed{\bf{s = 200 \: m }}}$

${\mathfrak{\underline{\purple{\:\:\:Additional \:Information:-\:\:\:}}}}$

☢ Equations Of Motion

$\boxed{</p><p></p><p>\begin{minipage}{3 cm} \\</p><p></p><p>\sf{\:\bullet\:\:v = u +at} \\ \\</p><p></p><p>\sf{\:\bullet\:\:s = ut + \frac{1}{2}\:at^{2} }\\ \\</p><p></p><p>\sf{\:\bullet\:\:v^{2} = u^{2} + 2as}\\ \\</p><p>\sf{\:\bullet\:\:s = \dfrac{1}{2} (u + v)t}\\ </p><p></p><p>\end{minipage}</p><p></p><p>}$

$\sf{Where,}$

$\:\:\:\:\bullet\:\:\:\textsf{v = Final velocity}$

$\:\:\:\:\bullet\:\:\:\textsf{u = Initial velocity}$

$\:\:\:\:\bullet\:\:\:\textsf{a = Acceleration}$

$\:\:\:\:\bullet\:\:\:\textsf{s = Distance}$

$\:\:\:\:\bullet\:\:\:\textsf{t = Time taken}$

Answer 2

Answer

• The distance covered will be 200 m

Explanation

Given

• Initial Velocity = 40 m/s
• Final Velocity = 0 m/s
• Times = 10 sec

To Find

• Distance covered during this time

Solution

● We shall first use the first equation of motion to find the acceleration of the body and then use the second (or) the third equation of motion to find the distance covered

✭ Acceleration of the vehicle

→ v = u+at

→ 0 = 40 + a×10

→ 0-40 = 10a

→ -40 = 10a

→ -40/10 = a

→ Acceleration = -4 m/s²

✭ Distance Covered

[Using second equation of motion]

→ s = ut+½at²

→ s = 40×10 + ½ × (-4) × 10²

→ s = 400 + -2×100

→ s = 400 + -200

→ s = 200 m

[Using third equation of motion]

→ v²-u² = 2as

→ 0²-40² = 2 × (-4) × s

→ 0-1600 = -8s

→ -1600/-8 = s

→ s = 200 m