A particle starts from the origin at t = 0 s with a velocity of 10.0 i ^ m/s and moves with a constant acceleration of (8 i + 2 j) m/s^2. (a) At what time is the x-coordinate of th... A particle starts from the origin at t = 0 s with a velocity of 10.0 i ^ m/s and moves with a constant acceleration of (8 i + 2 j) m/s^2. (a) At what time is the x-coordinate of the particle 16 m? (b) What is the speed of the particle at that time? What is the y-coordinate of the particle at that time? Read more

Steps to Solve

1. Understand the motion of the particle The particle has an initial velocity of $10.0 \hat{i} , \text{m/s}$ and an acceleration of $(8 \hat{i} + 2 \hat{j}) , \text{m/s}^2$. We will break down the motion in terms of the x and y coordinates separately.

2. Find the time at which the x-coordinate is 16 m We can use the equation for the x-coordinate:

$$ x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 $$

Substituting the values:

• $x_0 = 0 , \text{m}$
• $v_{0x} = 10.0 , \text{m/s}$
• $a_x = 8 , \text{m/s}^2$
• $x = 16 , \text{m}$

The equation becomes:

$$ 16 = 0 + 10.0 t + \frac{1}{2} (8) t^2 $$

This simplifies to:

$$ 16 = 10.0 t + 4 t^2 $$

Rearranging gives:

$$ 4 t^2 + 10 t - 16 = 0 $$

3. Solve the quadratic equation Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=4$, $b=10$, and $c=-16$:

Calculate the discriminant:

$$ b^2 - 4ac = 10^2 - 4(4)(-16) = 100 + 256 = 356 $$

Now, substitute into the quadratic formula:

$$ t = \frac{-10 \pm \sqrt{356}}{2(4)} $$

4. Calculate the roots of the equation The positive root gives the time $t$:

$$ t = \frac{-10 + \sqrt{356}}{8} $$

Calculating $\sqrt{356} \approx 18.87$, we have:

$$ t \approx \frac{-10 + 18.87}{8} \approx 1.11 , \text{s} $$

5. Find the speed of the particle at that time The speed can be found by calculating the velocity vectors in x and y at $t = 1.11 , \text{s}$.

The velocity in the x-direction:

$$ v_x = v_{0x} + a_x t = 10.0 + 8(1.11) = 10.0 + 8.88 \approx 18.88 , \text{m/s} $$

The velocity in the y-direction:

$$ v_y = v_{0y} + a_y t = 0 + 2(1.11) = 2.22 , \text{m/s} $$

Thus, the speed is given by:

$$ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(18.88)^2 + (2.22)^2} $$

6. Calculate the total speed Calculating the speed:

$$ v \approx \sqrt{356.67 + 4.93} \approx \sqrt{361.6} \approx 19.00 , \text{m/s} $$

7. Find the y-coordinate at that time Using the equation for y:

$$ y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 $$

Substituting in our values:

$$ y = 0 + 0 + \frac{1}{2}(2)(1.11)^2 \approx 1.23 , \text{m} $$