Question

The velocity of a particle moving along a straight line with variable acceleration as V is equal to 9tsquare + 4t+ 1 metre per second t in seconds the displacement of the particle during t is equal to zero to t is equal to 3

Answer 1

$Given:\\v(t)=9t^2+4t+1$

We need to find displacement of particle from t=0 to t=3

Integrate the equation,

dv=ds/dt

$\int\limits^s_0 {} \, ds = \int\limits^3_0 {v} \, dt \\\\s=\int\limits^3_0 {(9t^2+4t+1)} \, dt \\s=\frac{9t^3}3+\frac{4t^2}{2} +t \\s=3t^3+2t^2+t \left \{ {{t=3} \atop {t=0}} \right. \\=3(27)+2(9)+3\\\\=81+18+3\\=102m$