# A scooter moving at a speed of 10 m/s is stopped by applying breaks which produce a uniform acceleration of -0.5m/s*.how much distance will be covered by

the scooter before it stops

2
by Agu1

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the scooter before it stops

2
by Agu1

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final velocity......................=0 m/s

acceleration of ..............=-0.5 m/s²

apply formula v²-u²=2as

-100=2x-0.5xs

-100=-s

s=100

distance traveled by scooter before it stops is 100m

u=10m/s

a=-0.5m/s2 t=v-u/a

v=0 t=-10/-0.5=50s

S=ut+1/2at2

=500- 125

=675m