Question

Due to the small coefficient of friction, a car can't move on a ground covered with ice with an acceleration exceeding a = 0.5m/s². According to the rules of a competition, the car must go from point A to point B, which is located at right angles to the initial velocity of the car (figure), as quickly as possible. What is the minimum time needed for this if the distance AB = 375m and the initial velocity of the cart v = 10m/s? What is the car's trajectory? Mark correct options for trajectory and minimum time.

• A. Circle
• B. Parabola
• C. 40sec
• D. 50sec

Answer 1

Answer: (B), (D)

Parabola, 50sec

Answer 2

The problem asks us to find the minimum time required for a car to travel from point A to point B and to determine the trajectory of the car.

1. Set up the coordinate system and initial conditions: Let point A be the origin (0,0). The initial velocity v = 10 m/s is given. Let's assume it's along the positive x-axis. So, v_x(0) = 10 m/s and v_y(0) = 0. Point B is located at right angles to the initial velocity, at a distance AB = 375 m. This means point B is at (0, 375 m). The maximum allowed acceleration is a_max = 0.5 m/s². To minimize time, the car must use this maximum acceleration. Let the components of acceleration be a_x and a_y. The constraint is a_x^2 + a_y^2 = a_max^2 = (0.5)^2 = 0.25.

2. Equations of motion: The position of the car at time t is given by: x(t) = x(0) + v_x(0)t + (1/2)a_x t^2 y(t) = y(0) + v_y(0)t + (1/2)a_y t^2

Substituting initial conditions: x(t) = 0 + 10t + (1/2)a_x t^2 y(t) = 0 + 0t + (1/2)a_y t^2

At time t, the car reaches point B, so x(t) = 0 and y(t) = 375 m. From x(t) = 0: 0 = 10t + (1/2)a_x t^2 Since t > 0, we can divide by t: 0 = 10 + (1/2)a_x t a_x = -20/t (Equation 1)

From y(t) = 375: 375 = (1/2)a_y t^2 a_y = 750/t^2 (Equation 2)

3. Apply the acceleration constraint: Substitute a_x and a_y into a_x^2 + a_y^2 = 0.25: (-20/t)^2 + (750/t^2)^2 = 0.25 400/t^2 + 562500/t^4 = 0.25

Multiply the entire equation by t^4 to clear denominators: 400t^2 + 562500 = 0.25t^4

Rearrange into a quadratic equation in t^2: 0.25t^4 - 400t^2 - 562500 = 0 Multiply by 4 to simplify coefficients: t^4 - 1600t^2 - 2250000 = 0

Let X = t^2. The equation becomes: X^2 - 1600X - 2250000 = 0

Solve for X using the quadratic formula X = [-b ± sqrt(b^2 - 4ac)] / 2a: X = [1600 ± sqrt((-1600)^2 - 4 * 1 * (-2250000))] / (2 * 1) X = [1600 ± sqrt(2560000 + 9000000)] / 2 X = [1600 ± sqrt(11560000)] / 2

Calculate sqrt(11560000): sqrt(1156 * 10^4) = sqrt(1156) * sqrt(10^4) = 34 * 100 = 3400

So, X = [1600 ± 3400] / 2

Two possible values for X: X_1 = (1600 + 3400) / 2 = 5000 / 2 = 2500 X_2 = (1600 - 3400) / 2 = -1800 / 2 = -900

Since X = t^2, it must be non-negative. Therefore, X = 2500. t^2 = 2500 t = sqrt(2500) = 50 seconds (time must be positive).

The minimum time needed is 50 seconds.

4. Determine the car's trajectory: Now that we have t = 50 s, we can find the constant acceleration components: a_x = -20/t = -20/50 = -0.4 m/s^2 a_y = 750/t^2 = 750/(50^2) = 750/2500 = 0.3 m/s^2

The equations of motion are: x(t) = 10t + (1/2)(-0.4)t^2 = 10t - 0.2t^2 y(t) = (1/2)(0.3)t^2 = 0.15t^2

From the equation for y(t), we can express t^2 in terms of y: t^2 = y / 0.15 And t = sqrt(y / 0.15) = sqrt(20y/3) (since t > 0).

Substitute these into the equation for x(t): x = 10 * sqrt(20y/3) - 0.2 * (y/0.15) x = 10 * sqrt(20y/3) - (4/3)y

Rearrange the equation: x + (4/3)y = 10 * sqrt(20y/3) Square both sides: (x + (4/3)y)^2 = (10 * sqrt(20y/3))^2 (x + (4/3)y)^2 = 100 * (20y/3) (x + (4/3)y)^2 = (2000/3)y

This equation is of the form (Ax + By + C)^2 = D(Ex + Fy + G). This is the general equation of a parabola. The trajectory of a particle moving under constant acceleration (which is the case here, as a_x and a_y are constant) is always a parabola, unless the initial velocity is parallel to the acceleration (which is not the case here, as v(0) = (10,0) and a = (-0.4, 0.3)).

Thus, the car's trajectory is a parabola.

5. Conclusion: The minimum time needed is 50 seconds. The car's trajectory is a parabola.

Comparing with the given options: (a) Circle - Incorrect (b) Parabola - Correct (c) 40sec - Incorrect (d) 50sec - Correct

The correct options are (b) and (d).