v=9t^2+2t+9 m/s. if the body starts from x=3m, find the position at t=2s
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Answer:
The Body will be 48 m away from the initial point.
Explanation:
Given,
Initial velocity (u) = 0 m/s
Final Velocity (v) = (9t²+2t+9) m/s
Replacing the value of t by 2, we get,
= [9(2)²+(2*2)+9] m/s
= 36+4+9 m/s
= 49 m/s
Time (t) = 2s
Acceleration (a) = v = u+at
= 49 = 2a
= a = 24.5 m/s²
Distance (S) = ut+0.5at²
= 2+0.5[24.5*2*2]
= 2+0.5[98]
= 2+49
= 51 m
But, as the body starts from a distance of 3 m away from the initial position,
It'll distance of = (51-3) m
= 48 m
Hope this helps.
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