Final answer:
a)The magnitude of acceleration of the particle, when its speed is constant at 0.6 m/s, is 0.9 m/s².
b) The magnitude of acceleration of the particle, when its speed is 0.6 m/s and increasing at the rate of 1.2 m/s each second, is approximately 1.5 m/s².
Step-by-step explanation:
To calculate the magnitude of acceleration (a) of a particle moving in a circular path with a radius of 0.4 m, we can use the following formulas:
(a) If the speed is constant at 0.6 m/s:
The acceleration of a particle moving in a circular path at a constant speed is given by the formula:
a = v² / r
where v is the speed and r is the radius.
Substituting the given values, we have:
a = (0.6 m/s)² / 0.4 m
a = 0.9 m/s²
Therefore, the magnitude of acceleration (a) of the particle, when its speed is constant at 0.6 m/s, is 0.9 m/s².
(b) If the speed is 0.6 m/s and increasing at the rate of 1.2 m/s each second:
In this case, we need to consider the tangential acceleration (at) and the radial acceleration (ar).
The tangential acceleration (at) is the rate of change of speed and is given by the formula:
at = dv / dt
where dv is the change in velocity and dt is the change in time.
The radial acceleration (ar) is the centripetal acceleration and is given by the formula:
ar = v² / r
where v is the speed and r is the radius.
To calculate the total acceleration (a), we add the tangential acceleration (at) and the radial acceleration (ar):
a = √(at² + ar²)
In this case, the change in velocity (dv) is equal to 1.2 m/s and the change in time (dt) is equal to 1 second.
Substituting the given values, we have:
at = 1.2 m/s
ar = (0.6 m/s)² / 0.4 m = 0.9 m/s²
a = √((1.2 m/s)² + (0.9 m/s²)²)
a ≈ 1.5 m/s²
Therefore, the magnitude of acceleration (a) of the particle, when its speed is 0.6 m/s and increasing at the rate of 1.2 m/s each second, is approximately 1.5 m/s².