Asked by srikar
2.The distance travelled by a body moving with uniform acceleration in the nth second is given by Sn=3.8+0.4n,find the velocity of the body
Answers
Answered by
choco
Sn=3.8+0.4-----(eqn 1)
Sn=u+a(n-1/2)
=u+an-a/2
Sn=(u-a/2)+an------(eqn 2)
from 1 and 2
a=0.4m/s^2
u-a/2=3.8
u-0.4/2=3.8
u-0.2=3.8
u=6.0m/s
Sn=u+a(n-1/2)
=u+an-a/2
Sn=(u-a/2)+an------(eqn 2)
from 1 and 2
a=0.4m/s^2
u-a/2=3.8
u-0.4/2=3.8
u-0.2=3.8
u=6.0m/s
Answered by
Niharikaa Nayana
Good
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