Question

v=9t^2+2t+9 m/s. if the body starts from x=3m, find the position at t=2s​

Answer 1

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.