At t = 0, an automobile traveling north begins to make a turn. It follows one-quarter of the arc of a circle of radius 10.1 m until, at t = 1.52 s, it is traveling east. The car does not alter its speed during the turn.
(a) Find the car's speed.
(b) Find the change in its velocity during the turn.
(c) Find its average acceleration during the turn.
I already determined the speed for (a) to be 10.4 m/s.
I don't understand how to find the change in velocity (b). The equation in the book is v=wt. This didn't work so I would appreciate it if someone could help me understand this conceptually! Thanks!
To find the change in velocity during the turn, we need to determine the initial and final velocities of the car and then take the difference between them.
Given that the car follows one-quarter of the arc of a circle of radius 10.1 m, we know that the displacement of the car is equal to one-fourth of the circumference of the circle, which is (1/4)(2πr) = (1/4)(2π10.1 m) = 5.05π m.
We also know that the displacement is equal to the change in position, which is equal to the change in distance traveled by the car during the turn. Since the car does not alter its speed during the turn, it means that the magnitude of the initial velocity (v_i) is equal to the magnitude of the final velocity (v_f).
Let's assume that the initial velocity is directed north, and the final velocity is directed east. We can break down the initial and final velocities into their northward (V_n) and eastward (V_e) components.
At t = 0 (initially), the car is traveling north, so the initial velocity can be expressed as:
v_i = V_n (north)
At t = 1.52 s (finally), the car is traveling east, so the final velocity can be expressed as:
v_f = V_e (east)
Since the magnitude of the initial and final velocities is the same, we have:
|v_i| = |v_f|
Thus,
√(V_n^2) = √(V_e^2)
Simplifying, we have:
V_n = V_e
This means that the northward and eastward components of the velocity are equal in magnitude.
Now, recall that displacement (S) is defined as the change in position or distance traveled. It can be calculated using the formula:
S = v_avg * t
where v_avg is the average velocity during the time interval t. In this case, the car traveled one-fourth of the circumference of the circle during a time period of 1.52 s.
Now, we can determine the average velocity during the turn:
v_avg = S / t = (5.05π m) / (1.52 s) = 3.32π m/s
Since the average velocity during the turn is equal to the magnitude of both the initial and final velocities, we can express it in terms of the northward and eastward components:
v_avg = √(V_n^2 + V_e^2)
Since V_n = V_e, we can substitute it into the equation:
3.32π m/s = √(V_n^2 + V_n^2) = √(2V_n^2)
Squaring both sides, we have:
(3.32π m/s)^2 = 2V_n^2
Solving for V_n, we get:
V_n^2 = (3.32π m/s)^2 / 2
V_n = √[(3.32π m/s)^2 / 2] ≈ 3.27 m/s
Therefore, the northward component of the velocity is approximately 3.27 m/s.
Since the initial velocity is directed north and the final velocity is directed east, the change in velocity during the turn can be calculated using vector subtraction:
Δv = v_f - v_i
The magnitudes of the initial and final velocities are the same, so the magnitudes of the northward and eastward components are also the same:
|v_i| = √(V_n^2) = √(3.27 m/s)^2 ≈ 3.27 m/s
|v_f| = √(V_e^2) = √(3.27 m/s)^2 ≈ 3.27 m/s
Therefore, the change in velocity can be calculated as:
Δv = √((V_e - V_n)^2) = √((3.27 m/s)^2 + (3.27 m/s)^2) ≈ 4.62 m/s
So, the change in velocity during the turn is approximately 4.62 m/s.
Note: Make sure to double-check the calculations as rounding-off errors may occur.
To find the change in velocity during the turn, we need to consider the initial and final velocities of the car.
First, let's determine the initial velocity of the car. At t = 0, the car is traveling north, so its initial velocity is solely in the y-direction. Since the car does not alter its speed during the turn, the magnitude of the initial velocity is the same as the magnitude of the final velocity, which we found to be 10.4 m/s in part (a). Therefore, the initial velocity of the car is 10.4 m/s in the negative y-direction.
Next, let's determine the final velocity of the car. At t = 1.52 s, the car is traveling east, so its final velocity is solely in the x-direction. Since the car follows one-quarter of the arc of a circle, its path is circular and perpendicular to the final velocity. The radius of the circle is given as 10.1 m.
To calculate the final velocity, we can use the relationship between linear velocity and angular velocity. The angular velocity, represented by the symbol "w," can be calculated by dividing the angle covered by the car in radians (π/2, representing one-quarter of a full circle) by the time taken (1.52 s):
w = (π/2) / 1.52 s
Once we have the angular velocity, we can multiply it by the radius to find the final velocity:
v = w * r
Therefore, the final velocity of the car is:
v = [(π/2) / 1.52 s] * 10.1 m = 10.448 m/s
Now, to find the change in velocity, we can subtract the initial velocity from the final velocity:
Δv = v_final - v_initial
Δv = 10.448 m/s - (-10.4 m/s) = 20.848 m/s
So, the change in velocity during the turn is 20.848 m/s.
It's worth noting that the equation v = wt you mentioned is a simplified form of the relationship between linear velocity (v), angular velocity (w), and radius (r) for objects moving in a circle. In this particular scenario, we need to consider the radius of the circle and the time taken to determine the change in velocity accurately.