Math / Calculus

QuestionA particle moves along the x -axis with an initial velocity $\mathrm{v}_{\mathrm{x}}=50 \mathrm{ft} / \mathrm{sec}$ at the origin when $t=0$ . For the first 4 seconds it has no acceleration, and thereafter it is acted on by a retarding force which gives it a constant acceleration $\mathrm{ax}_{\mathrm{x}}=-10 \mathrm{ft} / \mathrm{sec}^{2}$ . Calculate the velocity and the $x$ -coordinate of the particle for the conditions of $t=8 \mathrm{sec}$ and $t=12 \mathrm{sec}$ and find the maximum positive $x$ -coordinate reached by the particle.

Studdy Solution

STEP 1

1. The particle moves along the x-axis.
2. Initial velocity $v_x = 50 \, \text{ft/s}$ at $t = 0$ .
3. No acceleration for the first 4 seconds.
4. A constant acceleration $a_x = -10 \, \text{ft/s}^2$ acts after 4 seconds.
5. We need to find the velocity and x-coordinate at $t = 8 \, \text{sec}$ and $t = 12 \, \text{sec}$ .
6. We need to find the maximum positive x-coordinate.

STEP 2

1. Determine the velocity and position for $t = 8 \, \text{sec}$ .
2. Determine the velocity and position for $t = 12 \, \text{sec}$ .
3. Find the maximum positive x-coordinate.

STEP 3

For $t = 8 \, \text{sec}$ :
1.1 Calculate the velocity at $t = 8 \, \text{sec}$ .
- For the first 4 seconds, the velocity remains constant: $v_x(4) = 50 \, \text{ft/s}$
- From $t = 4 \, \text{sec}$ to $t = 8 \, \text{sec}$ , use the equation: $v_x(t) = v_x(4) + a_x \cdot (t - 4)$ $v_x(8) = 50 + (-10) \cdot (8 - 4)$ $v_x(8) = 50 - 40 = 10 \, \text{ft/s}$
1.2 Calculate the position at $t = 8 \, \text{sec}$ .
- From $t = 0$ to $t = 4$ : $x(4) = v_x \cdot t = 50 \cdot 4 = 200 \, \text{ft}$
- From $t = 4$ to $t = 8$ : $x(8) = x(4) + v_x(4) \cdot (t - 4) + \frac{1}{2} a_x \cdot (t - 4)^2$ $x(8) = 200 + 50 \cdot 4 + \frac{1}{2} \cdot (-10) \cdot 4^2$ $x(8) = 200 + 200 - 80 = 320 \, \text{ft}$

STEP 4

For $t = 12 \, \text{sec}$ :
2.1 Calculate the velocity at $t = 12 \, \text{sec}$ .
- Use the equation: $v_x(12) = v_x(4) + a_x \cdot (12 - 4)$ $v_x(12) = 50 + (-10) \cdot (12 - 4)$ $v_x(12) = 50 - 80 = -30 \, \text{ft/s}$
2.2 Calculate the position at $t = 12 \, \text{sec}$ .
- From $t = 4$ to $t = 12$ : $x(12) = x(4) + v_x(4) \cdot (12 - 4) + \frac{1}{2} a_x \cdot (12 - 4)^2$ $x(12) = 200 + 50 \cdot 8 + \frac{1}{2} \cdot (-10) \cdot 8^2$ $x(12) = 200 + 400 - 320 = 280 \, \text{ft}$

STEP 5

Find the maximum positive x-coordinate.
- The maximum x-coordinate occurs when the velocity becomes zero. - Set $v_x(t) = 0$ and solve for $t$ : $0 = 50 + (-10) \cdot (t - 4)$ $10(t - 4) = 50$ $t - 4 = 5$ $t = 9 \, \text{sec}$
- Calculate the position at $t = 9 \, \text{sec}$ : $x(9) = x(4) + v_x(4) \cdot (9 - 4) + \frac{1}{2} a_x \cdot (9 - 4)^2$ $x(9) = 200 + 50 \cdot 5 + \frac{1}{2} \cdot (-10) \cdot 5^2$ $x(9) = 200 + 250 - 125 = 325 \, \text{ft}$
The velocity and position at $t = 8 \, \text{sec}$ are $10 \, \text{ft/s}$ and $320 \, \text{ft}$ , respectively. The velocity and position at $t = 12 \, \text{sec}$ are $-30 \, \text{ft/s}$ and $280 \, \text{ft}$ , respectively. The maximum positive x-coordinate reached by the particle is $325 \, \text{ft}$ .

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