Question

1.The velocity of a particle moving on the x -axis is given by v = x^2 + x, where 'x' is in m and 'v' is in m/s
What is the position (in meters) when it's acceleration is 30 m/(s^2)?

2. A boy standing on the top of a tower of height 54 ft. throws a packet with a speed of 20 ft./s directly
aiming for his friend who is standing 72 ft. away from the foot of the tower. The packet falls short of the friend on the ground by X*16/3. The value X is _______ .

Answer 1

Answer on Question #44668, Physics, Mechanics | Kinematics | Dynamics

1. The velocity of a particle moving on the $x$-axis is given by $v = x^2 + x$, where 'x' is in m and 'v' is in m/s

What is the position (in meters) when it's acceleration is $30 \, \text{m/(s}^2)$?

2. A boy standing on the top of a tower of height 54 ft. throws a packet with a speed of 20 ft./s directly

aiming for his friend who is standing 72 ft. away from the foot of the tower. The packet falls short of the friend on the ground by $x^16/3$. The value $x$ is ________.

Solution:

#1

$a_1 = 30 \frac{\text{m}}{\text{s}}$ – final acceleration of the particle;

$x_1$ – position when it’s acceleration $a_1$;

The velocity of a particle moving on the $x$-axis:

$$v = x^2 + x$$

The acceleration of the particle is $\left(v = \frac{dx}{dt}\right)$

$$a = v'(t) = \frac{d(x^2 + x)}{dt} = 2x \cdot \frac{dx}{dt} + \frac{dx}{dt} = 2x \cdot v + v$$

(1) in (2):

$$a = 2x \cdot (x^2 + x) + x^2 + x = 2x^3 + 2x^2 + x^2 + x = 2x^3 + 3x^2 + x$$

For the acceleration $a_1 = 30 \frac{\text{m}}{\text{s}}$:

$$a_1 = 2x_1^3 + 3x_1^2 + x_1$$ $$2x_1^3 + 3x_1^2 + x_1 - 30 = 0$$

Real solution of the equation:

$$x_1 = 2$$

Answer: the position (in meters) when it's acceleration is $30 \frac{\text{m}}{\text{s}}$ is $2\text{m}$.

$\mathrm{h} = 54$ ft.-height of the tower;

$\mathrm{V} = 20\frac{\mathrm{ft}}{\mathrm{s}}$ - initial velocity of the packet;

$\mathrm{d} = 72$ ft.-distance from the tower to friend;

$\mathrm{s} = \mathrm{X}\cdot \frac{16}{3}$ -distance from packet to friend;

$\alpha$ - angle between vertical and direction of the motion;

Components of the velocity along X-axis and Y-axis:

$\mathrm{V_x = V\cos\alpha;V_y = V\sin\alpha;}$

From the right triangle:

$\sin \alpha = \frac{\mathrm{d}}{\mathrm{AB}} = \frac{\mathrm{d}}{\sqrt{\mathrm{d}^2 + \mathrm{h}^2}} = \frac{72\mathrm{ft}}{\sqrt{(72\mathrm{ft})^2 + (54\mathrm{ft.})^2}} = \frac{4}{5}$

$\cos \alpha = \frac{\mathrm{h}}{\mathrm{AB}} = \frac{\mathrm{h}}{\sqrt{\mathrm{d}^2 + \mathrm{h}^2}} = \frac{54\mathrm{ft}}{\sqrt{(72\mathrm{ft})^2 + (54\mathrm{ft.})^2}} = \frac{3}{5}$

$\tan \alpha = \frac{\mathrm{d}}{\mathrm{h}} = \frac{72\mathrm{ft.}}{54\mathrm{ft.}} = \frac{4}{3}$

Equation of motion of the particle along X-axis:

$\mathrm{x:s = V_x t = Vt\sin\alpha}$

$$t = \frac {s}{V \sin \alpha} \quad (1)$$

Equation of motion of the particle along Y-axis ($g = 32\frac{\mathrm{ft}}{\mathrm{s}^2}$):

$$y: h = V t \cos \alpha + \frac {g t ^ {2}}{2} \quad (2)$$

(1) in (2):

$$h = V \frac {s}{V \sin \alpha} \cos \alpha + \frac {g \left(\frac {s}{V \sin \alpha}\right) ^ {2}}{2}$$ $$h = s \tan \alpha + \frac {g s ^ {2}}{2 V ^ {2} \sin^ {2} \alpha}$$ $$54 = \frac {4}{3} \cdot \frac {16}{3} X + \frac {32 \frac {\mathrm {ft}}{\mathrm {s} ^ {2}} \cdot \left(\frac {16}{3} X\right) ^ {2}}{2 \cdot \left(20 \frac {\mathrm {ft}}{\mathrm {s}}\right) ^ {2} \cdot \left(\frac {4}{5}\right) ^ {2}}$$ $$54 = \frac {16}{9} X (X + 4)$$ $$\frac {16}{9} X ^ {2} + \frac {64}{9} X - 54 = 0$$

Solutions of the quadratic equation:

$$X _ {1} = \frac {1}{4} \left(- 8 - 5 \sqrt {22}\right) \approx - 7.86$$ $$X _ {2} = \frac {1}{4} (5 \sqrt {22} - 8) \approx 3.86$$

We need only positive root, hence $X = 3.86$.

**Answer**: The value X is 3.86

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