Question

A car travelling at speed 72km/hr suddenly applies the brake with the deceleration of 5m/s2. Find the stopping distance of the car​

(plz tell the answer quickly)

Answer 1

Answer:

40 meters..

Explanation:

as per the formula... v^2=u^2 – 2 a s

or,2as=u^

or,s=(u^2)/(2a)

Answer 2

${\huge{\purple{\bold{Hey Friend...}}}}$

Here is the Solution..

${\red{Initial\:Speed, \: u}}$ = 72 km/hr = 20 m/s

${\red{Final\: speed ,\: v}}$ = 0

Let ${\red{Stopping \:Distance}}$ = 's'

${\red{Deceleration/Negative\: acceleration}}$ = -5m/s²

By Applying Second Equation of Motion

⇒⇒ ${\boxed{\huge{\blue{\bold{v^2 = u^2 + 2as}}}}}$

→ v² = u² - 2as

→ 0 = 20² + 2 x -5 x s

→ 0 = 400 - 10s

→ 10s = 400

→ s = 40 m

∴ Stopping Distance = 40 m

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${\boxed{\huge{\red{Ans = 40 m}}}}$

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I Hope This Will Help You