Question

Car is running at 72km/h is slowed down to 18km/h by the application of breaks over a distance of 20m. Find the deceleration of car​

Answer 1

Answer:

• $\sf{-9.375 \:m/s^2}$

Explanation:

${\underline{\underline{\sf{\red{{{\star}}\:\:\:Given:-}}}}}$

$\tiny\:\:\:\:\:\:\:\:\bullet\:\:\:\sf\orange{Initial\: velocity \:(u) = 72km/h = \dfrac{72\times5 }{18} = 20m/s}\\\tiny\:\:\:\:\:\:\:\:\bullet\:\:\:\sf\green{Final \ velocity \ (v) = 18km/h = \dfrac{18\times5}{18} = 5m/s}\\\tiny\:\:\:\:\:\:\:\:\bullet\:\:\:\sf\blue{Distance \ covered \ by \ car \ (s) = 20 m}$

${\underline{\underline{\sf{\orange{{\star}\:\:\:ToFind:-}}}}}$

$\tiny\:\:\:\:\:\:\:\:\bullet\:\:\:\sf\purple{Deceleration \ of \ car (a)}$

${\underline{\underline{\sf{\blue{{\star}\:\:\: Solution:-}}}}}$

$\dag\:\underline{\mathfrak{\purple{Using\: 3rd\: equation \:of \:motion }}}$

$\red{\underline{\boxed{\sf{v^2 = u^2 + 2as}}}}$

${\underline{\sf{\:\:\:\:Where,\:\:\:}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\orange{v = final \:velocity}\\ \tiny\:\:\:\:\bullet\:\:\:\sf\green{u = initial \:velocity}\\\tiny \:\:\:\:\bullet\:\:\:\sf\blue{a = acceleration\: or \:deceleration}\\\tiny \:\:\:\:\bullet\:\:\:\sf\red{s = distance}$

$\dag\:\underline{\mathfrak{\purple{Substituting \: the \: values}}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ 5^2 = 20^2 + 2\times a \times 20}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ 25 = 400 + 40a}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ 25 -400 = 40a}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ -375 = 40a}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ \dfrac{-375}{40} = a}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ a = -9.375 \:m/s^2}$

${\underline{\sf{\:\:\:\:Deceleration \ of \ the \ car \ is \ 9.375 \:m/s^{2}\:\:\:}}}$