Question

A particle experiences constant acceleration for 20 s after starting from rest. If it travels a distance $1 in the first 10 second and a distance S2 in the next 10 second, then (a) S2 = 2S1 (b) S2 = 3S1 (c) S2 = 4S1 (d) S2 = 5S1​

Answer 1

Answer:

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