一个质点沿直线运动，通过固定点 $O$ 之后 $t$ 秒时，其速度 $v$ m/s 由 $v = 22 + 7t - 2t^2$ 给出。质点在 $P$ 处瞬时静止。求：

1. Find the expression for acceleration by differentiating velocity

$a = \frac{dv}{dt} = 7 - 4t$

2. Find the time when acceleration is zero

$7 - 4t = 0$
$4t = 7$
$t = 1.75$

3. Calculate the velocity at this time

$v = 22 + 7(1.75) - 2(1.75)^2$
$v = 22 + 12.25 - 6.125$
$v = 28.125$

1. Set velocity to zero for instantaneous rest

At instantaneous rest, $v = 0$.
$22 + 7t - 2t^2 = 0$
$2t^2 - 7t - 22 = 0$

2. Solve the quadratic equation for $t$

$(2t - 11)(t + 2) = 0$
$t = 5.5$ or $t = -2$
Since $t \ge 0$, $t = 5.5$.

1. Find the value of $t$ at $P$ (from part ii)

At $P$, $v = 0$.
$22 + 7t - 2t^2 = 0$
$(11 - 2t)(2 + t) = 0$
$t = 5.5$ or $t = -2$ (reject)

2. Set up the definite integral for displacement

Distance $OP = \int_0^{5.5} v \, dt$
$= \int_0^{5.5} (22 + 7t - 2t^2) \, dt$

3. Integrate the velocity function

$= \left[ 22t + \frac{7t^2}{2} - \frac{2t^3}{3} \right]_0^{5.5}$

4. Evaluate the definite integral

$= \left( 22(5.5) + \frac{7(5.5)^2}{2} - \frac{2(5.5)^3}{3} \right) - 0$
$= 121 + 105.875 - 110.9167$
$= 115.9583$
$= 115\frac{23}{24}$ m