Question

4. A scooter moving at a speed of 10 m/s is stopped by applying brakes which produce a uniform acceleration
of, -0.5 m/s2. How much distance will be covered by the scooter before it stops?
​

Answer 1

Here, u = 10 m/s

v = 0 m/s

retardation, a = -0.5 m/s^2

Using the third equation of motion,

v^2 = u^2 + 2as

=> 0 = 10^2 + 2(-0.5)(s)

=> 100 - s = 0

=> s = 100 m

Thus, the scooter travels a distance of 100 maters before it stops.

Answer 2

Answer :

${\red\bigstar\Large\bold{\underline{\textsf{\green{Given:}}}}}$

• Initial Velocity : 10 m/s

• Final Velocity : 0 ( as brakes are applied )

• Acceleration : -0.5 $\sf {m}^{2}$ ( retardation )

${\red\bigstar\Large\bold{\underline{\textsf{\green{To\:Find:}}}}}$

• Distance covered before the scooter stops

${\red\bigstar\Large\bold{\underline{\textsf{\green{Solution:}}}}}$

☯ FORMULA USED :

$\large\bold{\boxed{\sf{\pink{{v}^{2}\:-\:{u}^{2}\:=\:2as}}}}$

➡ $\sf 0\:-\:{10}^{2}\:=\:2(-\:0.5)s$

➡ $\sf -100\:=\:-s$

➡ $\sf s\:=\:100$

${\orange\bigstar\Large\bold{\underline{\textsf{\red{Distance\:travelled\:=\:100\:m}}}}}$

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