Velocity of a body moving with uniform
acceleration of 3 m/s is changed through 30 m/
sin certain time. Average velocity ofbody during
this time is 30 m/s. Distance covered by it during
this time is???
Answers
Answered by
1
Answer:
V
avg
=
t
D
⇒0.34=
t
3.06
⇒t=
0.34
3.06
= 9 sec
We have, v=u+at
v−u=at
0.18=a×9
⇒a=
9
0.18
=0.02 m/s
2
Explanation:
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