An Uniformly accelerated object covers 6.5m during 2nd second and 8.5m during 4th second of its motion.Find its initial velocity as well as uniform acceleration
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Answered by
3
Answer:
initial velocity is equal to 5m/s and uniform acceleration in this case is 1 m/s^2
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Answered by
2
Answer:
the distance travelled in nth second
S
n
th
=u+
2
1
a(2n−1)
When n=5 then s
n
th
=65m
∴65=u+
2
1
a(2×5−1)
or 65=u+
2
9
az
or u+
2
9
a=65.....(i)
When n=9 then s
n
th
=105m
∴1005=u+
2
1
a(2×9−1)
or 105=u+
2
1
a(2×9−1)
or 105=u+
2
17
a
or u+
2
17
a=105......(2)
Subtracting equation (1) from equation (2) we get ,
2
8
a=40⇒a=
4
40
=10m/s
or a=10m/s
Substituting the value of a in equation (1) we have
u+
2
9
×10=65⇒u+45=65
∴u=65−45=20ms
−1
or u=20ms
−1
s
20
th
=u+
2
1
a(2×20−1)=20+
2
1
×10(40−1)
=20+5×39=20+195=215m
or s
20
th
=215m
Total distance travelled in 20s
s=ut+
2
1
at
2
=20×20+
2
1
×10×20×20
=400+5××400=400+2000
=2400m
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