a partical covers 100 m of distance in first 10 second of its motion and 200 m of disance and next 15 second. with constant acceleration of the particle also find distence covered and next 5 second?
Answers
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
Answered By
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Explanation:
ANSWER
It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion
Case I
s=ut+
2
1
at
2
10=5u+
2
1
a(5)
2
20=10u+25a
4=2u+5a..............(1)
Case 2
In next 3s the particle covers more 10m distance. So
20=8u+
2
1
a(8)
2
5=2u+8a.........(2)
On solving equation (1) and (2)
4=2u+5a
5=2u+8a
a=
3
1
m/s
2
Put the value of a in equation (1)
u=
6
7
m/s
Now to find distance in next 10 s. total time will be 10s
s=
6
7
×10+
2
1
×
3
1
×(10)
2
s=28.33m
Distance travelled in next 2 sec
s=28.33−20=8.33m