Question

A scooter moving at a speed of 10m/s is stopped by applying brake produce a uniform acceleration of -0.5m/s*s .how much distance will be covered by the scooter before its stops.

Answer 1

initial velocity u=10m/s
retardation a=-0.5m/s²
final velocity v=0 (brake is applied)
let the distance travel be S
using Newton law of motion,
v²-u²=2aS
S=v²-u²/2a
S=0-(10)²/2(-0.5)
S=-100/-1
S=100m
Distance travel by the scooter before it stopped is 100 metres.

Answer 2

QuEsTiOn :-

A scooter moving at a speed of 10 m/s is stopped by applying brakes which produce a uniform acceleration of 0.5 m/s, how much distance covered by scooter before it stops ?

AnSweR :-

• The distance covered by the scooter is 100 metres

Given :-

• Initial speed ( u ) = 10 m/s

• Final speed ( v ) = 0 m/s

• Acceleration ( a ) = -5 m/s²

To be calculated :-

• Calculate the distance covered by the scooter before it stops$.$

Solution :-

According to the third equation of motion,

• v² = u² + 2as

Where ,

Initial speed = ( u )

Final speed = ( v )

Acceleration = ( a )

Distance covered = ( s)

Now ,

➤ Putting the values in the third equation of motion

we get :

⇒ ( 0 )² = ( 10 )² + 2 × ( -0.5 ) × s

⇒ 0 = 100 - s

⇒ s = 100

⇒ s = 100/10

⇒ s = 100 m

Thus,

The distance covered by the scooter is 100 metres.

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➤ Extra shots :-

• v = u + at ( first equation of motion )

• s = ut + ½ at² ( second equation of motion )

• v² = u² + 2as ( third equation of motion )

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