A car starts from rest on a curve with a radius of 150 and accelerates at 1.2.Through what angle will the car have traveled when the magnitude of its total acceleration is 2.2 ?
Given:R=150m,
tangential component of acceleration a(τ)=1.2 m/s²,
a=2.2 m/s²,
φ=?
a²=a(n)²+a(τ)²,
where a(n) is the normal( centripetal) component of acceleration.
a(n)= sqrt(a²-a(τ)²) = sqrt(2.2²-1.2²) = 1.84 m/s²
a(n)=ω²•R =>
ω² = a(n)/R
The angular acceleration ε= a(τ)/R.
From
φ=ε•t²/2
ω=ε•t,
we obtain
φ=ω²/2•ε= a(n)/2•a(τ)=
=1.84/2•1.2=0.77 rad.
To find the angle through which the car will have traveled when the magnitude of its total acceleration is 2.2, we can use the concept of centripetal acceleration.
First, we need to calculate the magnitude of the centripetal acceleration using the formula:
ac = v^2 / r
where ac is the centripetal acceleration, v is the velocity of the car, and r is the radius of the curve.
Given that the car starts from rest, the initial velocity (v0) is 0 m/s. We can calculate the final velocity (v) using the formula:
v^2 = v0^2 + 2ad
where a is the acceleration and d is the distance traveled. In this case, the distance traveled is the arc length of the curve.
Let's assume that the car travels through an angle θ when the magnitude of its total acceleration is 2.2 m/s^2. Using the relationship between the arc length (s) and the radius (r), we can calculate the distance traveled:
s = rθ
Now, we need to find the final velocity of the car. Rearranging the formula for final velocity (v), we get:
v = √(v0^2 + 2ad)
Since the initial velocity is 0, the formula simplifies to:
v = √(2ad)
Substituting the formula for distance traveled (s) into the equation, we get:
v = √(2a(rθ))
Now, we can calculate the centripetal acceleration (ac):
ac = v^2 / r = (2a(rθ))^2 / r
Simplifying, we have:
ac = 4a^2θ
Given that the magnitude of the total acceleration is 2.2 m/s^2, we can set up an equation:
2.2 = 4a^2θ
Solving for θ, we get:
θ = 2.2 / (4a^2)
Substituting the given values of a (1.2 m/s^2) and solving, we have:
θ = 2.2 / (4 * (1.2)^2)
Evaluating this expression, we find:
θ ≈ 0.384 radians
Therefore, the car will have traveled approximately 0.384 radians when the magnitude of its total acceleration is 2.2 m/s^2.