Question

a car moving at a speed of 10m/s it stopped by appling bricks which produce a uniform acceleration of -0.5 m/s² how much distance will be covered by the car before it stops?​

Answer 1

$\bigstar\rm\blue{GIVEN}\bigstar$

• $\rm\blue{Initial\:Velocity=10m/s}$

• $\rm\red{Final\:Velocity=0m/s}$

• $\rm\pink{Acceleration=-0.5m/s^{2}}$

$\bigstar\rm\blue{TO\:FIND}\bigstar$

• The Distance Covered by that car before it stops.

$\bigstar\rm\blue{FORMULAE\:USED}\bigstar$

• ${\boxed{\rm{\red{v^2-u^2=2as}}}}$

Where,

V= Final Velocity

u= Initial Velocity

a=Acceleration

S = Distance.

Now, Using the formula

$\implies\rm\blue{v^2-U^2=2as}$

$\implies\rm\blue{(0)^2-10^2=2\times{0.5}\times{S}}$

100=1×S

$\implies{100=S}$

Thus, The distance covered by the car is 100m.

$\bigstar\rm{MORE\:TO\:KNOW}\bigstar$

• The retardation is the Acceleration at opposite direction.

• $\rm\blue{V=U+at}$

Answer 2

$\huge\tt{Answer:-}$

$\bf{Given:-}$

• Initial Velocity $(u)$ = $10m \ {s}^{-1}$
• Final Velocity $(v)$ = $0m \ {s}^{-1}$
• Acceleration $(a)$ = $-0.5 m \ {s}^{-2}$

$\bf{To \ Find:-}$

Distance $(s)$ it will cover before it will stop.

________________...

We know,

$2as = {v}^{2} - {u}^{2}$

$\implies s = \frac{ {v}^{2} - {u}^{2}}{2a}$

$\implies = \frac{- {u}^{2}}{2a}$

After putting the values :-

$s = \frac{- {(10m \ {s}^{-1})}^{2}}{2(-0.5 m \ {s}^{-2})}$

$\implies s = \frac{- (100 \ {m}^{2} \ {s}^{-2})}{-1 m \ {s}^{-1}}$

$\implies s = \frac{\cancel{- 100 \ {m}^{2} \ {s}^{-2}}}{\cancel{-1 m \ {s}^{-1}}}$

$\implies s = 100 m$ $...(Ans.)$

________________...

Formula Used here:-

• $2as = {v}^{2} - {u}^{2}$

Certain Formulae that can help you:-

• $s = ut + \frac{1}{2}a{t}^{2}$ (☑ This formula will also work if some other quantities are given. )
• $v = u + at$