Question

A scooter moving with constant accelaration cover 60 m in 6s between two point. It's speed is 15m/s when it passes second point then at what prior distance from the first point the scooter was at rest

Answer 1

Answer:

7.5m

Explanation:

between 1st and 2nd point,

v=15m/s

v=15m/ss=60m

v=15m/ss=60mt=6s

v=u+at

a=(v-u)/t

S=ut + 1/2at^2

S=ut + 1/2 × (v-u)/t × t^2

60=6u + 1/2 × (15-u)/6 × 6×6

60=6u + 3(15-u)

60=6u + 45 - 3u

60-45=3u

u=5 m/s

a=(15-5)/6=10/6 m/s^2

between 1st point and starting point,

v=5 m/s

v=5 m/sa=10/6 m/s^2

v=5 m/sa=10/6 m/s^2u=0

v=u+at

t=(v-u)/a

as u=0

S=1/2 × a × t^2

S=1/2 × a × (v-u)/a × (v-u)/a

=1/2 × 10/6 × 5/(10/6) × 5/(10/6)

=1/2 × 10/6 × 30/10 × 30/10

=1/2 × 10/6 × 3 × 3

=1/2 × 15

=7.5 m