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A particle experiences constant acceleration for 20 seconds after starting from rest. It it travels a distance D1 in the first 10 seconds and distance D2 in next 10 seconds then :-
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(a) D1 = D2
(b) D2 = 2D1
(c) D2 = 3D1
(d) D2 = 4D1
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Answer with explanation! :)
Answers
Given :-
A particle experiences constant acceleration for 20 seconds after starting from rest. It it travels a distance D1 in the first 10 seconds and distance D2 in next 10 seconds
To Find :-
Relation between distances
Solution :-
We know that
s = ut + 1/2at²
In case of D1
⇒ d1 = 0(t) + 1/2 × a × (10)²
⇒ d1 = 0 + 1/2 × a × 100
⇒ d1 = 0 + 50a
⇒ d1 = 50a
Now
⇒ v = u + at
⇒ v = 0 + 10(a)
⇒ v = 0 + 10a
⇒ v = 10a
Final velocity of D1 = Initial velocity of D2
In case of D2
Using the same formula
⇒ d2 = (10a)(10) + 1/2 × a × (10)²
⇒ d2 = 100a + 1/2 × a × 100
⇒ d2 = 100a + 50a
⇒ d2 = 150a
Now
d1 = 50a
d2 = 150a
On dividing both
d1/d2 = 50/150
d1/d2 = 1/3
3d1 = d2
d2 = 3d1 Or,
D2 = 3D1
Option C is correct
[tex][/tex]
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