A car starts from rest on a curve with a radius of 150m and tangential acceleration of 1.0m/s2 .

Through what angle will the car have traveled when the magnitude of its total acceleration is 3.0m/s2 ?

To determine the angle through which the car will have traveled, we need to use the concept of centripetal acceleration and tangential acceleration.

Centripetal acceleration (ac) is the acceleration experienced by an object moving in a circular path. It is given by the formula:

ac = v^2 / r

Where:
ac = centripetal acceleration
v = velocity of the object
r = radius of the circular path

In this scenario, the car is starting from rest, so its initial velocity (v0) is zero. The tangential acceleration (at) is given as 1.0 m/s^2 and the radius (r) is 150 m.

We can rearrange the centripetal acceleration formula to solve for the velocity (v):

v = √(ac * r)

Now, we need to find the velocity of the car when its magnitude of total acceleration is 3.0 m/s^2. To do this, we can use the Pythagorean theorem, as the total acceleration (a) can be represented as the vector sum of the centripetal acceleration (ac) and the tangential acceleration (at):

a = √(ac^2 + at^2)

We can rearrange this equation to solve for ac:

ac = √(a^2 - at^2)

Given that a = 3.0 m/s^2 and at = 1.0 m/s^2.

Now, we can substitute the known values into the equations:

ac = √(3.0^2 - 1.0^2)
ac = √(9.0 - 1.0)
ac = √8.0
ac ≈ 2.83 m/s^2

Using the equation to find the velocity:

v = √(ac * r)
v = √(2.83 * 150)
v = √424.5
v ≈ 20.6 m/s

Next, we can find the time taken (t) for the car to reach this velocity using the equation:

v = v0 + at

Since the car starts from rest (v0 = 0), we can rearrange the equation to solve for t:

t = v / a

Substituting the known values:

t = 20.6 / 1.0
t = 20.6 s

Now, to find the angle (θ) through which the car will have traveled, we can use the equation:

θ = ω0 * t + (1/2) * α * t^2

Where:
θ = angle in radians
ω0 = initial angular velocity (0, since the car starts from rest)
α = angular acceleration (ac / r)
t = time (20.6 s)

Substituting the values:

θ = 0 * 20.6 + (1/2) * (2.83 / 150) * (20.6^2)
θ = 0 + (1/2) * (0.01886) * (425.16)
θ = 0 + 4.0025
θ ≈ 4.0025 radians

Therefore, the car will have traveled approximately 4.0025 radians when the magnitude of its total acceleration is 3.0 m/s^2.