A car starts from rest on a curve with a radius of 150 m and tangential acceleration of 1.5 m/s2
Through what angle will the car have traveled when the magnitude of its total acceleration is 3.0 m/s2 ?
A loudspeaker of mass 16.0 kg is suspended a distance of h = 2.00 m below the ceiling by two cables that make equal angles with the ceiling. Each cable has a length of l = 2.90 m
What is the tension T in each of the cables?
Use 9.80 m/s2 for the magnitude of the acceleration due to gravity.
To find the angle through which the car will have traveled, we need to use the concept of centripetal acceleration and tangential acceleration.
First, let's find the centripetal acceleration (ac). The centripetal acceleration is the acceleration toward the center of the curve and is given by the formula:
ac = v^2 / r
where v is the velocity and r is the radius of the curve.
Since the car starts from rest, the initial velocity (v0) is 0. So, the velocity (v) can be calculated using the formula of tangential acceleration (at):
at = change in velocity / time
In this case, the change in velocity is v - v0, which is just v since v0 is 0. So, we have:
at = v / t
Rearranging the formula, we get:
v = at * t
Substituting this value of v in the formula for centripetal acceleration, we have:
ac = (at * t)^2 / r
Now, let's solve these equations to find the time (t) for which the magnitude of the total acceleration (atotal) is 3.0 m/s^2.
We know that the total acceleration can be calculated using the Pythagorean theorem:
atotal = sqrt(ac^2 + at^2)
Substituting the values we know, we have:
3.0 m/s^2 = sqrt((at * t)^2 / r^2 + at^2)
Square both sides of the equation to get rid of the square root:
9.0 m^2/s^4 = (at * t)^2 / r^2 + at^2
Expand the equation:
9.0 m^2/s^4 = (at^2 * t^2) / r^2 + at^2
Multiply both sides by r^2 to eliminate the fraction:
9.0 m^2/s^4 * r^2 = at^2 * t^2 + at^2 * r^2
Simplify:
9.0 m^2/s^2 * r^2 = at^2 * (t^2 + r^2)
Divide both sides by at^2:
9.0 m^2/s^2 * r^2 / at^2 = t^2 + r^2
Rearrange the equation:
t^2 = 9.0 m^2/s^2 * r^2 / at^2 - r^2
t^2 = (9.0 m^2/s^2 / at^2) * r^2 - r^2
Now, we can solve for t by substituting the given values: at = 1.5 m/s^2 and r = 150 m.
t^2 = (9.0 m^2/s^2 / (1.5 m/s^2)^2) * (150 m)^2 - (150 m)^2
Simplify:
t^2 = (9.0 m^2/s^2 / 2.25 m^2/s^2) * 22,500 m^2 - 22,500 m^2
t^2 = 4 * 22,500 m^2 - 22,500 m^2
t^2 = 90,000 m^2 - 22,500 m^2
t^2 = 67,500 m^2
t = sqrt(67,500 m^2) ≈ 259.81 m
Now that we know the time (t), we can find the angle (θ) using the formula:
θ = at * t
Substituting the given values, we have:
θ = 1.5 m/s^2 * 259.81 s ≈ 389.71 radians
Therefore, the car will have traveled approximately 389.71 radians when the magnitude of its total acceleration is 3.0 m/s^2.