(i) To find the velocity of the particle at point A, we need to integrate the acceleration function to get the velocity function. The acceleration $ a $ is given by:

$$ a = -2 - 2t $$

The velocity $ v $ is the integral of the acceleration with respect to time $ t $:

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

Integrating term by term:

$$ v = \int -2 \, dt + \int -2t \, dt $$
$$ v = -2t - t^2 + C $$

where $ C $ is the constant of integration. We need to determine $ C $ using the given information. The particle comes to rest at $ t = 4 $, so $ v(4) = 0 $:

$$ 0 = -2(4) - (4)^2 + C $$
$$ 0 = -8 - 16 + C $$
$$ C = 24 $$

Thus, the velocity function is:

$$ v = -2t - t^2 + 24 $$

To find the velocity at point A (when $ t = 0 $):

$$ v(0) = -2(0) - (0)^2 + 24 $$
$$ v(0) = 24 \, \text{ms}^{-1} $$

(ii) To find the distance AB, we need to integrate the velocity function to get the position function. The velocity $ v $ is given by:

$$ v = -2t - t^2 + 24 $$

The position $ s $ is the integral of the velocity with respect to time $ t $:

$$ s = \int v \, dt = \int (-2t - t^2 + 24) \, dt $$

Integrating term by term:

$$ s = \int -2t \, dt + \int -t^2 \, dt + \int 24 \, dt $$
$$ s = -t^2 - \frac{t^3}{3} + 24t + D $$

where $ D $ is the constant of integration. We can set $ D = 0 $ since we are measuring the distance from point A (where $ s = 0 $ when $ t = 0 $):

$$ s = -t^2 - \frac{t^3}{3} + 24t $$

To find the distance AB, we evaluate $ s $ at $ t = 4 $:

$$ s(4) = -4^2 - \frac{4^3}{3} + 24(4) $$
$$ s(4) = -16 - \frac{64}{3} + 96 $$
$$ s(4) = -16 - 21.33 + 96 $$
$$ s(4) = 58.67 \, \text{m} $$

Therefore, the distance AB is approximately $ 58.67 \, \text{m} $.

Question

(i) To find the velocity of the particle at point A, we need to integrate the acceleration function to get the velocity function. The acceleration $ a $ is given by:

$$ a = -2 - 2t $$

The velocity $ v $ is the integral of the acceleration with respect to time $ t $:

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

Integrating term by term:

$$ v = \int -2 \, dt + \int -2t \, dt $$
$$ v = -2t - t^2 + C $$

where $ C $ is the constant of integration. We need to determine $ C $ using the given information. The particle comes to rest at $ t = 4 $, so $ v(4) = 0 $:

$$ 0 = -2(4) - (4)^2 + C $$
$$ 0 = -8 - 16 + C $$
$$ C = 24 $$

Thus, the velocity function is:

$$ v = -2t - t^2 + 24 $$

To find the velocity at point A (when $ t = 0 $):

$$ v(0) = -2(0) - (0)^2 + 24 $$
$$ v(0) = 24 \, \text{ms}^{-1} $$

(ii) To find the distance AB, we need to integrate the velocity function to get the position function. The velocity $ v $ is given by:

$$ v = -2t - t^2 + 24 $$

The position $ s $ is the integral of the velocity with respect to time $ t $:

$$ s = \int v \, dt = \int (-2t - t^2 + 24) \, dt $$

Integrating term by term:

$$ s = \int -2t \, dt + \int -t^2 \, dt + \int 24 \, dt $$
$$ s = -t^2 - \frac{t^3}{3} + 24t + D $$

where $ D $ is the constant of integration. We can set $ D = 0 $ since we are measuring the distance from point A (where $ s = 0 $ when $ t = 0 $):

$$ s = -t^2 - \frac{t^3}{3} + 24t $$

To find the distance AB, we evaluate $ s $ at $ t = 4 $:

$$ s(4) = -4^2 - \frac{4^3}{3} + 24(4) $$
$$ s(4) = -16 - \frac{64}{3} + 96 $$
$$ s(4) = -16 - 21.33 + 96 $$
$$ s(4) = 58.67 \, \text{m} $$

Therefore, the distance AB is approximately $ 58.67 \, \text{m} $.

Knowee AI · Accepted Answer

(i) To find the velocity of the particle at point A, we need to integrate the acceleration function to get the velocity function. The acceleration $ a $ is given by:

$$ a = -2 - 2t $$

The velocity $ v $ is the integral of the acceleration with respect to time $ t $:

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

Integrating term by term:

$$ v = \int -2 \, dt + \int -2t \, dt $$
$$ v = -2t - t^2 + C $$

where $ C $ is the constant of integration. We need to determine $ C $ using the given information. The particle comes to rest at $ t = 4 $, so $ v(4) = 0 $:

$$ 0 = -2(4) - (4)^2 + C $$
$$ 0 = -8 - 16 + C $$
$$ C = 24 $$

Thus, the velocity function is:

$$ v = -2t - t^2 + 24 $$

To find the velocity at point A (when $ t = 0 $):

$$ v(0) = -2(0) - (0)^2 + 24 $$
$$ v(0) = 24 \, \text{ms}^{-1} $$

(ii) To find the distance AB, we need to integrate the velocity function to get the position function. The velocity $ v $ is given by:

$$ v = -2t - t^2 + 24 $$

The position $ s $ is the integral of the velocity with respect to time $ t $:

$$ s = \int v \, dt = \int (-2t - t^2 + 24) \, dt $$

Integrating term by term:

$$ s = \int -2t \, dt + \int -t^2 \, dt + \int 24 \, dt $$
$$ s = -t^2 - \frac{t^3}{3} + 24t + D $$

where $ D $ is the constant of integration. We can set $ D = 0 $ since we are measuring the distance from point A (where $ s = 0 $ when $ t = 0 $):

$$ s = -t^2 - \frac{t^3}{3} + 24t $$

To find the distance AB, we evaluate $ s $ at $ t = 4 $:

$$ s(4) = -4^2 - \frac{4^3}{3} + 24(4) $$
$$ s(4) = -16 - \frac{64}{3} + 96 $$
$$ s(4) = -16 - 21.33 + 96 $$
$$ s(4) = 58.67 \, \text{m} $$

Therefore, the distance AB is approximately $ 58.67 \, \text{m} $.

11 A particle travels in a straight line so that, t seconds after passing a fixed point A on the line, itsacceleration, a ms–2, is given by a = –2 – 2t. It comes to rest at a point B when t = 4.(i) Find the velocity of the particle at A. [4](ii) Find the distance AB.

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