Question

A particle moving in a straight line experience constant acceleration for 20s after starting from rest. If it travels a distance S1 in the first 10s and distance S2 in the next 10s,then find relation between S1 and S2.​

Answer 1

Explanation:

Let a be the constant acceleration of the particle. Thens=ut+

2

1

at

2

or s

1

=0+

2

1

×a×(10)

2

=50aand s

2

=[0+

2

1

a(20)

2

]−50a=150a

∴s

2

=3s

1

Alternatively:Let a be constant acceleration and

s=ut+

2

1

at

2

, then s

1

=0+

2

1

×a×100=50a

Velocity after 10 sec. is v=0+10a

So, s

2

=10a×10+

2

1

a×100=150a⇒s

2

=3s

1

Let a be constant acceleration, using s=ut+

2

1

at

2

so distance coreved in first 10 seconds s

1

=0+

2

1

×a×100=50a

Velocity after 10 sec. is v=0+10a

So, distance covered in next 10 seconds s

2

=10a×10+

2

1

a×100=150a⇒s

2

=3s

1