To find the velocity of the particle at point A, we need to integrate the acceleration function to get the velocity function. The given acceleration is $ a = -2 - 2t $.

Step 1: Integrate the acceleration function to find the velocity function.
$$ a = \frac{dv}{dt} = -2 - 2t $$

Integrate both sides with respect to $ t $:
$$ \int \frac{dv}{dt} \, dt = \int (-2 - 2t) \, dt $$

$$ v(t) = \int (-2 - 2t) \, dt $$

$$ v(t) = -2t - t^2 + C $$

Step 2: Determine the constant of integration $ C $.

We know that the particle comes to rest at $ t = 4 $, so $ v(4) = 0 $:
$$ 0 = -2(4) - (4)^2 + C $$

$$ 0 = -8 - 16 + C $$

$$ 0 = -24 + C $$

$$ C = 24 $$

Step 3: Write the velocity function with the determined constant.
$$ v(t) = -2t - t^2 + 24 $$

Step 4: Find the velocity at point A, which is when $ t = 0 $.
$$ v(0) = -2(0) - (0)^2 + 24 $$

$$ v(0) = 24 $$

Therefore, the velocity of the particle at point A is $ 24 \, \text{ms}^{-1} $.

Question

To find the velocity of the particle at point A, we need to integrate the acceleration function to get the velocity function. The given acceleration is $ a = -2 - 2t $.

Step 1: Integrate the acceleration function to find the velocity function.
$$ a = \frac{dv}{dt} = -2 - 2t $$

Integrate both sides with respect to $ t $:
$$ \int \frac{dv}{dt} \, dt = \int (-2 - 2t) \, dt $$

$$ v(t) = \int (-2 - 2t) \, dt $$

$$ v(t) = -2t - t^2 + C $$

Step 2: Determine the constant of integration $ C $.

We know that the particle comes to rest at $ t = 4 $, so $ v(4) = 0 $:
$$ 0 = -2(4) - (4)^2 + C $$

$$ 0 = -8 - 16 + C $$

$$ 0 = -24 + C $$

$$ C = 24 $$

Step 3: Write the velocity function with the determined constant.
$$ v(t) = -2t - t^2 + 24 $$

Step 4: Find the velocity at point A, which is when $ t = 0 $.
$$ v(0) = -2(0) - (0)^2 + 24 $$

$$ v(0) = 24 $$

Therefore, the velocity of the particle at point A is $ 24 \, \text{ms}^{-1} $.

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