Question

A particle moving at a velocity of 4.6 m/s in the positive x direction is given an acceleration of 5.3 m/s² in the positive y direction for 6.6 s. What is the final speed of the particle?

Answer 1

To find the final speed of the particle, we can use the equation vf = sqrt(vi^2 + 2*a*d), where vi is the initial velocity, a is the acceleration, and d is the distance. Plugging in the given values, we find that the final speed of the particle is approximately 232.06 m/s.

Step-by-step explanation:

To find the final speed of the particle, we can use the equation:

vf = sqrt(vi^2 + 2*a*d)

where:

• vf is the final velocity (which is what we're trying to find)
• vi is the initial velocity, which is 4.6 m/s
• a is the acceleration, which is 5.3 m/s²
• d is the distance, which we'll calculate using the formula d = vi*t + 0.5*a*t^2
• t is the time, which is given as 6.6 s

Plugging in the values, we get:

d = 4.6 m/s * 6.6 s + 0.5 * 5.3 m/s² * (6.6 s)^2

d = 30.36 m + 0.5 * 5.3 m/s² * 43.56 s²

d = 30.36 m + 0.5 * 5.3 m/s² * 1903.44 s²

d = 30.36 m + 5039.08 m = 5069.44 m

Now we can substitute the values of vi, a, and d into the equation for vf:

vf = sqrt((4.6 m/s)^2 + 2 * 5.3 m/s² * 5069.44 m)

vf = sqrt(21.16 m²/s² + 53692.352 m²/s²)

vf ≈ 232.06 m/s

So, the final speed of the particle is approximately 232.06 m/s.