A particle revolves in a horizontal circle of radius 3.20 m. At a particular instant, its
acceleration is 1.20 m/s
2
in a direction that makes an angle of 55.0º to its direction of motion.
Determine its speed (a) at this moment, and (b) 1.50 s later, assuming constant magnitude of
tangential acceleration.
I am not certain how the 55 degrees is measured? Is it in the plane of revolution? towards inside, or outside?
55 from the tangent, so i think its accelerating.
Centripetal acceleration equals tangent 55 times 1.05. Obtain the velocity at that point with that. V=Vinitial+at, where a is the anet in the direction of motion. Therefore, get the original velocity, find the anet through a cosine with the tangent acceleration, and solve the kinematics equation.
To determine the speed of the particle at the given instant, we can use the equation for centripetal acceleration:
a = (v^2) / r
where:
a = centripetal acceleration
v = speed of the particle
r = radius of the circle
In this case, the given acceleration is 1.20 m/s^2. We can substitute these values into the equation and solve for v:
1.20 m/s^2 = (v^2) / 3.20 m
To solve for v, we rearrange the equation:
v^2 = 1.20 m/s^2 * 3.20 m
v^2 = 3.84 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 1.96 m/s
So, the speed of the particle at this moment is approximately 1.96 m/s.
To determine the speed of the particle 1.50 seconds later, we can use the equation for tangential acceleration:
a = Δv / Δt
where:
a = tangential acceleration
Δv = change in velocity
Δt = change in time
In this case, the tangential acceleration is constant, so we can assume the change in velocity is proportional to the change in time. Let's assume Δv is the change in velocity and Δt is the change in time (1.50 s).
We can write the equation as:
1.20 m/s^2 = Δv / 1.50 s
To solve for Δv, we rearrange the equation:
Δv = 1.20 m/s^2 * 1.50 s
Δv = 1.80 m/s
Since the speed of the particle is the magnitude of its velocity, and the direction of the velocity remains the same, the speed 1.50 seconds later will be the initial speed plus the change in velocity:
v = 1.96 m/s + 1.80 m/s
v ≈ 3.76 m/s
So, the speed of the particle 1.50 seconds later will be approximately 3.76 m/s.