Question

A car moving along a straight highway with speed of 126 km/hr is broughtto astop within a distance of 200 m . What is the acceleration of the car (assumed uniform) and how long does it take for the car to stop.

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• A car moving with
• U=126km/h
• Stopping after few minutes applying the brake
• V=0
• S=200m

${\sf{\green{\underline{\large{To\:find}}}}}$

• Acceleration
• Time taken to stop

${\sf{\pink{\underline{\Large{Explanation}}}}}$

☞Changing the units in m or s

• U=126km/h = 126 × 5/18=35m/s

Now From equation of motion

V²=u²+2as

¶utting the value

0=35×35+2a×200

=>(-35×35)200=2a

=>49/16=a

=>a=3.05

=>~3m/s²

Now for time

V=u+at

=>-35=3t

=>t=11.6s

Answer 2

______________________________

✰Question ✰

⋄ A car moving along a straight highway with speed of 126 km/hr is broughtto astop within a distance of 200 m . What is the acceleration of the car (assumed uniform) and how long does it take for the car to stop.

_______________________________

❍ Given :-

⋄ Initial velocity of car (u) =126km/h=35m/s

⋄ Final velocity (v) = 0m/s

⋄ Distance covered by car (s) =200m

_______________________________

❍To Find :-

⋄ Acceleration of the car and the time taken by car to stop.

_______________________________

❍ Solution :-

By using using 3rd equation of motion

v² =u² +2as

0²= 35² +2× a × 200

400a = 1225

a=1225/400

a =3.06m/s²

∴The acceleration of the car is 3.06m/s²

And, now time taken by car to stop

v=u+at

0=35+3.06×t

t=35/3.06

t=11.43s

∴The time taken by car to stop is 11.43s