A particle at rest undergoes an acceleration of 2.5 m/s^2 to the right and 3.2 m/s^2 up. What is its speed after 5.9s? Answer in m/s
The magnitude of the acceleration is
a = sqrt[2.5^2 + 3.2^2] = 4.06 m/s^2,
since the two directions are perpendicular.
The speed after 5.9s is 5.9 a
You do the multiplication.
To find the speed of the particle after 5.9 seconds, we can use the equations of motion.
First, let's break down the given accelerations into their components. Since the acceleration to the right is 2.5 m/s² and the acceleration up is 3.2 m/s², we can represent these as ax = 2.5 m/s² and ay = 3.2 m/s².
Next, we can find the displacement in the horizontal direction (x-direction) and vertical direction (y-direction) using the following kinematic equations:
For x-direction:
s = ut + 0.5at²
Since the initial velocity (u) is zero (the particle is at rest), this simplifies to:
sx = 0.5axt²
Substituting the given values, we get:
sx = 0.5(2.5 m/s²)(5.9 s)² = 0.5(2.5)(34.81) = 43.51 meters
For y-direction:
sy = 0.5ayt²
Substituting the given values, we get:
sy = 0.5(3.2 m/s²)(5.9 s)² = 0.5(3.2)(34.81) = 57.41 meters
Now, we can calculate the resultant displacement (s) using the Pythagorean theorem:
s = √(sx² + sy²)
Substituting the calculated values, we get:
s = √(43.51² + 57.41²) = √(1891.4801 + 3297.8881) = √(5189.3682) = 72.06 meters
Finally, we can find the speed (v) using the equation:
v = s / t
Substituting the calculated values, we get:
v = 72.06 meters / 5.9 seconds ≈ 12.23 m/s
Therefore, the particle's speed after 5.9 seconds is approximately 12.23 m/s.