A particle's acceleration is given by a=(10−t) m/s². It starts from rest. If its velocity is 2 m/s when t=1 sec, find its velocity and displacement in terms of t.

Acceleration is the rate of change of velocity, so $a = \frac{dv}{dt}$.

$$v = \int a \, dt$$ $$v = \int (10-t) \, dt$$ $$v = 10t - \frac{t^2}{2} + C_1$$

We are given that $v = 2 \text{ m/s}$ when $t = 1 \text{ sec}$.

$$2 = 10(1) - \frac{1^2}{2} + C_1$$ $$2 = 10 - 0.5 + C_1$$ $$2 = 9.5 + C_1$$ $$C_1 = 2 - 9.5$$ $$C_1 = -7.5$$

$$v(t) = 10t - \frac{t^2}{2} - 7.5$$

The velocity in terms of $t$ is $\boxed{v(t) = 10t - \frac{t^2}{2} - 7.5 \text{ m/s}}$.

a) (ii) Its displacement after t seconds.

Velocity is the rate of change of displacement, so $v = \frac{ds}{dt}$.

$$s = \int v \, dt$$ $$s = \int \left(10t - \frac{t^2}{2} - 7.5\right) \, dt$$ $$s = 10\frac{t^2}{2} - \frac{1}{2}\frac{t^3}{3} - 7.5t + C_2$$ $$s(t) = 5t^2 - \frac{t^3}{6} - 7.5t + C_2$$

The particle "starts from rest", which implies that its initial displacement is 0 at $t=0$.

$$0 = 5(0)^2 - \frac{0^3}{6} - 7.5(0) + C_2$$ $$C_2 = 0$$

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