1. **State the problem:** A body moves along a straight line through point O. Its displacement at time $t$ seconds is given by $$s = t^3 + 4t^2 - 5t$$ We need to find: (a) The velocity $v$ at time $t$. (b) The acceleration at $t=2$ seconds. 2. **Recall formulas:** Velocity is the first derivative of displacement with respect to time: $$v = \frac{ds}{dt}$$ Acceleration is the derivative of velocity with respect to time: $$a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$$ 3. **Find velocity $v$:** Differentiate $s$: $$v = \frac{d}{dt}(t^3 + 4t^2 - 5t) = 3t^2 + 8t - 5$$ 4. **Find acceleration $a$:** Differentiate $v$: $$a = \frac{d}{dt}(3t^2 + 8t - 5) = 6t + 8$$ 5. **Calculate acceleration at $t=2$ seconds:** $$a = 6(2) + 8 = 12 + 8 = 20$$ **Final answers:** (a) Velocity: $v = 3t^2 + 8t - 5$ (b) Acceleration at 2 seconds: $20$ m/s$^2$