Answer to Question #180122 in Physics for Abd

Question

-A particle moving in a straight line has an acceleration of (2t – 3) 𝑚/〖sec〗^2 at time t seconds. The particle is initially 1m from O, a fixed point on the line, with a velocity of 2m/s. Find the time when the velocity is zero. Find the displacement of particle from O when t = 5sec?

Answer 1

The velocity is $v(t) = \int a(t) dt= 2\frac{t^2}{2} - 3 t + C = t^2 - 3 t + C$. Using initial condition, $v(0) = 2 = C$, from where $v(t) = t^2 - 3 t + 2$. Velocity is zero, when $v(t) = 0 = t^2 - 3 t +2$, which has two solutions $t = 1; t = 2$, hence the velocity is zero at these two times.

The coordinate of the particle is $x(t) = \int v(t) dt = \int (t^2 - 3 t + 2) dt = \frac{t^3}{3} - \frac{3}{2}t^2 + 2 t + C$. Using initial condition, $x(0) = 1 = C$, obtain $x(t) = \frac{t^3}{3} - \frac{3}{2} t^2 + 2 t +1$.

Displacement is $s(t) = x(t) - x(0) = \frac{t^3}{3} - \frac{3}{2} t^2 + 2 t$, and $s(5) = \frac{85}{6} m \approx 14.17 m$