At some instant, a particle traveling in a horizontal circular path of radius 7.80 m has a total acceleration with a magnitude of 15.0 m/s2 and a constant tangential acceleration of 12.0 m/s2. Determine the speed of the particle at this instant(initial speed) and (1/8) revolution later(final speed).

Why did the particle go on a circular path? Because it wanted to make a smooth curve! Now, let's calculate the initial speed and final speed.

The total acceleration of the particle can be broken into two components: radial and tangential acceleration. The radial acceleration is responsible for keeping the particle on the circular path, while the tangential acceleration changes the speed of the particle.

Given:
Radius (r) = 7.80 m
Total acceleration (a) = 15.0 m/s²
Tangential acceleration (at) = 12.0 m/s²
(1/8) revolution = 1/8 * 2π rad

To find the initial speed (vi), we need to remember that at the beginning, the tangential acceleration is zero because the particle starts from rest. So, we can say that the total acceleration at the start is equal to the radial acceleration, which can be obtained using the formula: ar = vi² / r.

Using this equation, we can rearrange it to solve for vi:

vi = sqrt(ar * r)

Now, let's substitute the values:

ar = a - at
ar = 15.0 m/s² - 12.0 m/s²
ar = 3.0 m/s²

vi = sqrt(3.0 m/s² * 7.80 m)
vi ≈ 7.17 m/s

So, the initial speed is approximately 7.17 m/s.

To find the final speed (vf) after (1/8) revolution later, we can use the formula:

vf = sqrt(vi² + 2 * at * (1/8 * 2π * r))

vf = sqrt(7.17 m/s * 7.17 m/s + 2 * 12.0 m/s² * (1/8 * 2π * 7.80 m))

vf ≈ sqrt(51.4 m²/s² + 2 * 12.0 m/s² * (1.57 * 7.80 m))

vf ≈ sqrt(51.4 m²/s² + 2 * 12.0 m/s² * 24.37 m)

vf ≈ sqrt(51.4 m²/s² + 587.66 m²/s²)

vf ≈ sqrt(639.06 m²/s²)

vf ≈ 25.3 m/s

So, the final speed (after 1/8 revolution) is approximately 25.3 m/s.

Remember, these values are approximate as we rounded off during the calculations.

To find the initial speed of the particle, we can use the equation for total acceleration:

a = √(at^2 + ar^2)

Where a is the total acceleration, at is the tangential acceleration, and ar is the radial (centripetal) acceleration.

Given:
a = 15.0 m/s^2
at = 12.0 m/s^2
r = 7.80 m

Using the equation, we can solve for ar:

ar = √(a^2 - at^2)
= √(15.0^2 - 12.0^2)
= √(225.0 - 144.0)
= √81.0
= 9.00 m/s^2

Now, let's find the initial speed:

v_initial = √(ar * r)
= √(9.00 * 7.80)
= √(70.20)
≈ 8.38 m/s

To find the final speed (after 1/8 revolution), we need to consider the change in angular displacement. 1/8 of a revolution is equivalent to an angular displacement of:

θ = (1/8) * 2π
= π/4 rad

The final speed can be calculated using the formula:

v_final = √(v_initial^2 + 2 * at * θ)

Substituting the given values:

v_final = √(8.38^2 + 2 * 12.0 * (π/4))
= √(70.2444 + 6π)
≈ 11.86 m/s

Therefore, the initial speed of the particle is approximately 8.38 m/s, and the final speed (after 1/8 revolution) is approximately 11.86 m/s.

To solve this problem, we need to understand the concept of circular motion and the relation between centripetal acceleration, tangential acceleration, and total acceleration.

In circular motion, an object travels along a circular path with a constant speed. It undergoes centripetal acceleration towards the center of the circle, which keeps it in the curved path. Additionally, it may have tangential acceleration, which is perpendicular to the centripetal acceleration and affects the object's speed.

In this question, we are given the magnitude of the total acceleration (15.0 m/s²) and the constant tangential acceleration (12.0 m/s²), and we need to find the speed of the particle at the initial instant and after (1/8) revolution.

To find the initial speed, we can use the fact that the total acceleration in circular motion is the vector sum of centripetal acceleration (aᵣ) and tangential acceleration (aₜ):

a = √(aᵣ² + aₜ²)

Given that a = 15.0 m/s² and aₜ = 12.0 m/s², we can calculate aᵣ:

aᵣ = √(a² - aₜ²) = √(15.0² - 12.0²) ≈ √(225 - 144) = √81 = 9.0 m/s²

Now, the initial speed (v₀) is related to the centripetal acceleration (aᵣ) by the equation:

aᵣ = v₀² / r

where r is the radius of the circular path. In this case, r = 7.80 m. Rearranging the equation:

v₀ = √(aᵣ * r) = √(9.0 * 7.80) ≈ √70.2 ≈ 8.38 m/s

Therefore, the initial speed of the particle is approximately 8.38 m/s.

To find the final speed after (1/8) revolution, we can use the equations of motion for circular motion. The total acceleration remains constant, so we can use the equation:

v² = v₀² + 2a * d

where v is the final speed, v₀ is the initial speed, a is the total acceleration, and d is the distance traveled. In this case, the particle travels (1/8) revolution, which corresponds to an angle of 45 degrees in the circular path. The distance traveled can be calculated as:

d = (θ/360) * 2πr

where θ is the angle in degrees and r is the radius. In this case, θ = 45 degrees and r = 7.80 m. Substituting the values into the equation:

d = (45/360) * 2π * 7.80 ≈ (0.125) * 2π * 7.80 ≈ 0.98π ≈ 3.08 m

Now, substituting the values into the equation for final speed:

v² = v₀² + 2a * d
v² = (8.38)² + 2 * 15.0 * 3.08
v² = 70.0844 + 92.4
v² ≈ 162.4844
v ≈ √162.4844 ≈ 12.75 m/s

Therefore, the final speed of the particle, after (1/8) revolution, is approximately 12.75 m/s.

total acceleration^2=centacc^2 + tangentialacc^2

solve for centripetal acceleration

then v^r/r=centripacc
solve for v.

Now for later,
wf^2=wi^2+2 alpha*displacement
alpha*r=tangential acceleration, displacement=1/8 *2PI

vf^2/r^2=v^2/r^2 + 12/r * 1/8 *2PI
solve for Vf