Question

A particle travels to the right at a constant
rate of 8.8 m/s. It suddenly is given a vertical
acceleration of 2.6 m/s2
for 3.8 s.
What is its direction of travel after the
acceleration with respect to the horizontal?
Answer between
−180◦
and +180◦ .
Answer in units
of◦
008 (part 2 of 2) 10.0 points
What is the speed at this time?
Answer in units of m/s
PS:I need this no later than 11/19/12 at 9:00p.m and i already did the first part.

Answer 1

Question#19095

A particle travels to the right at a constant rate of $8.8 \, \text{m/s}$. It suddenly is given a vertical acceleration of $2.6 \, \text{m/s}^2$ for $3.8 \, \text{s}$.

What is its direction of travel after the acceleration with respect to the horizontal?

Answer between $-180{}^{\circ}$ and $+180{}^{\circ}$. Answer in units of 008 (part 2 of 2) 10.0 points. What is the speed at this time? Answer in units of m/s

Solution:

Let:

$$v_x = 8.8 \, \text{m/s}$$ $$a = 2.6 \, \text{m/s}^2$$ $$t = 3.8 \, \text{s}$$

$v-?$, $a-?$

The result of due acceleration is velocity on horizontal direction:

$$v_y = a t$$

The angle of results velocity is:

$$\alpha = \operatorname{arctg} \frac{v_x}{v_y} = \operatorname{arctg} \frac{v_x}{a t}$$

The value of velocity is:

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{v_x^2 + a^2 t^2}$$ $$\alpha = \operatorname{arctg} \frac{8.8}{2.6 \cdot 3.8} = 41.69{}^{\circ} = 41{}^{\circ} 41'$$ $$v = \sqrt{8.8^2 + 2.6^2 \cdot 3.8^2} = 13.23 \, \text{m/s}$$

Answer: direction on angel $41{}^{\circ}41'$ (to x-axis), velocity $13.23 \, \text{m/s}$.