Question

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

Answer 1

Given:-

• Initial velocity,u = 10 m/s

• Final velocity,v = 0 m/s

• Acceleration,a = -0.5 m/s² [ retardation ]

To be calculated:-

Calculate the distance covered by scooter before it stops ?

Formula used:-

v² = u² + 2as

Solution:-

From third equation of motion,we have

v² = u² + 2as

★ Substituting the values in the above formula,we get :

⇒ ( 0 )² = ( 10 )² + 2 × -0.5 × s

⇒ 0 = 100 + 2 × - 5/10 × s

⇒ 0 = 100 - 1s

⇒ s = 100 m

Hence,the distance covered by scooter is 100 m.

Answer 2

Answer:

Given:

• A scooter moving at a speed of 10m/s is stopped by applying brakes which produce a uniform acceleration of -0.5 m per second square.

Find:

• How much distance will be covered by scooter before it stops ?

Know terms:

1. Initial velocity = (u)
2. Final velocity = (v)
3. Acceleration = (a)
4. Distance travelled = (s)

Calculations:

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

Calculations:

$\longrightarrow\bold{(0)^{2} = (10)^{2} + 2 \times -0.5 \times s}$

$\longrightarrow\bold{0 = 100 + 2 \times \dfrac{-5}{10} \times s}$

$\longrightarrow\bold{0 = 100 - 1}$

$\longrightarrow{\sf{\underline{\boxed{\red{\sf{100 \: m}}}}}}$

Therefore, the distance covered by scooter before it stops is 100 m.