A particle at rest undergoes an acceleration

of 2.9 m/s2 to the right and 2.6 m/s2 up.
What is its speed after 5.3 s?
Answer in units of m/s

hghgh

To find the speed of the particle after 5.3 seconds, we need to determine its final velocity using the given accelerations.

Since the particle is at rest initially, its initial velocity can be considered zero.

We can consider horizontal and vertical motions separately because they are independent of each other.

Horizontal Motion:
The acceleration in the horizontal direction is 2.9 m/s^2 to the right. Since the particle starts at rest, its horizontal velocity at any time t is equal to the product of acceleration and time: v_horizontal = a_horizontal * t. Therefore, v_horizontal = 2.9 m/s^2 * 5.3 s = 15.37 m/s to the right.

Vertical Motion:
The acceleration in the vertical direction is 2.6 m/s^2 upwards. Using the same formula, v_vertical = a_vertical * t = 2.6 m/s^2 * 5.3 s = 13.78 m/s upwards.

Now, to find the overall speed of the particle, we need to find the magnitude of its final velocity, combining the horizontal and vertical velocities using the Pythagorean theorem:

v_final = √(v_horizontal^2 + v_vertical^2)
v_final = √(15.37 m/s)^2 + (13.78 m/s)^2
v_final = √(235.8569 m^2/s^2 + 189.5684 m^2/s^2)
v_final = √(425.4253 m^2/s^2)
v_final ≈ 20.62 m/s

Therefore, the speed of the particle after 5.3 seconds is approximately 20.62 m/s.