Question

a car is moving on a straight road after covering a distance of 420m in 18s it turns back and stops after 8s half the way calculate average velocity of the car in first 20s​

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ average \ velocity \ is \ 23.775 \ m \ s^{-1}}$

$\sf{in \ first \ 20 \ s}$

$\sf\orange{Given:}$

$\sf{\leadsto{A \ car \ covers \ a \ distance \ of \ 420 \ m}}$

$\sf{in \ 18 \ s}$

$\sf{\leadsto{The \ car \ turns \ back \ and \ stops \ after}}$

$\sf{8 \ s \ half \ the \ way. }$

$\sf\pink{To \ find:}$

$\sf{Average \ velocity \ of \ the \ car \ in \ the \ first}$

$\sf{20 \ s}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Half \ distance=\dfrac{420}{2}=210 \ m}$

$\sf{The \ car \ covers \ half \ a \ distance \ in \ 8 \ s}$

$\sf{i.e. \ The \ car \ covers \ 210 \ m \ in \ 8 \ s}$

$\sf{Time \ \propto \ Distance}$

$\sf{The \ distance \ cover \ in \ 2 \ s}$

$\sf{=\dfrac{210}{4}=52.5 \ m}$

$\sf{Total \ distance \ cover \ in \ first \ 20 \ s}$

$\sf{=420+52.5=472.5 \ m}$

$\boxed{\sf{Average \ velocity=\dfrac{Total \ distance}{Time}}}$

$\sf{\mapsto{\therefore{Average \ velocity=\dfrac{475.5}{20}}}}$

$\sf{\mapsto{\therefore{Average \ velocity=23.775 \ m \ s^{-1}}}}$

$\sf\purple{\tt{\therefore{The \ average \ velocity \ is \ 23.775 \ m \ s^{-1}}}}$

$\sf\purple{\tt{in \ first \ 20 \ s}}$

Answer 2

Answer

The average velocity is 23.78 m/s (approx)