"In-Depth MCQs: Maximum Height, Range, and Time in Projectile Motion, Plus Circular Motion and Vectors"

"Expert MCQs: Projectile, Circular, Vectors"

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1. A projectile is thrown with an initial velocity of 20 m/s at an angle of 30° with the horizontal. What is the maximum height reached by the projectile? (Take \( g = 9.8 \, \text{m/s}^2 \))

A. 5 m

B. 7.5 m

C. 10 m

D. 15 m

The correct answer is 5 m. Given the initial velocity \( u = 20 \, \text{m/s} \) and angle of projection \( \theta = 30^\circ \), the formula for the maximum height \( H \) is: \[ H = \frac{u^2 \sin^2 \theta}{2g} \] Substituting the values, we get: \[ H = \frac{(20)^2 \sin^2 30^\circ}{2 \times 9.8} = \frac{400 \times (0.5)^2}{19.6} = \frac{100}{19.6} \approx 5.1 \, \text{m} \] Thus, the maximum height is approximately 5 m.

2. A projectile is launched with a velocity of 25 m/s at an angle of 45° to the horizontal. What is the range of the projectile? (Take \( g = 9.8 \, \text{m/s}^2 \))

A. 50 m

B. 55 m

C. 63.7 m

D. 70 m

The correct answer is 63.7 m. The formula for the range \( R \) of a projectile is: \[ R = \frac{u^2 \sin 2\theta}{g} \] Here, \( u = 25 \, \text{m/s} \) and \( \theta = 45^\circ \). Since \( \sin 90^\circ = 1 \): \[ R = \frac{(25)^2 \sin 90^\circ}{9.8} = \frac{625}{9.8} \approx 63.7 \, \text{m} \]

3. A ball is projected horizontally from the top of a tower with a speed of 15 m/s. If the time of flight is 3 s, what is the horizontal range of the projectile? (Take \( g = 9.8 \, \text{m/s}^2 \))

A. 30 m

B. 60 m

C. 45 m

D. 75 m

The correct answer is 60 m. The horizontal range \( R \) for a projectile launched horizontally is calculated using the formula: \[ R = u \times t \] Where \( u = 15 \, \text{m/s} \) and \( t = 3 \, \text{s} \): \[ R = 15 \times 3 = 45 \, \text{m} \]

4. A projectile is launched with an initial velocity of 40 m/s at an angle of 30° to the horizontal. What is the time of flight of the projectile? (Take \( g = 9.8 \, \text{m/s}^2 \))

A. 4.08 s

B. 5.10 s

C. 6.12 s

D. 7.15 s

The correct answer is 4.08 s. The time of flight (\( T \)) for a projectile is given by: \[ T = \frac{2u \sin \theta}{g} \] Substituting the values: \[ T = \frac{2 \times 40 \times \sin 30^\circ}{9.8} \] Since \( \sin 30^\circ = 0.5 \): \[ T = \frac{2 \times 40 \times 0.5}{9.8} = \frac{40}{9.8} \approx 4.08 \, \text{s} \]

5. A projectile is launched with an initial velocity of 20 m/s at an angle of 60° to the horizontal. What is the maximum height reached by the projectile? (Take \( g = 9.8 \, \text{m/s}^2 \))

A. 10.2 m

B. 12.5 m

C. 15.3 m

D. 18.0 m

The correct answer is 15.3 m. The maximum height (\( H \)) is given by: \[ H = \frac{u^2 \sin^2 \theta}{2g} \] Substituting the values: \[ H = \frac{20^2 \sin^2 60^\circ}{2 \times 9.8} \] Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \): \[ \sin^2 60^\circ \approx 0.750 \] \[ H = \frac{400 \times 0.750}{2 \times 9.8} = \frac{300}{19.6} \approx 15.3 \, \text{m} \]

6. A projectile is thrown with an initial velocity of 35 m/s at an angle of 45° with the horizontal. What is the time taken to reach the maximum height? (Take \( g = 9.8 \, \text{m/s}^2 \))

A. 2.55 s

B. 3.57 s

C. 4.08 s

D. 5.10 s

The correct answer is 2.55 s. The time to reach the maximum height (\( t_{\text{max}} \)) is given by: \[ t_{\text{max}} = \frac{u \sin \theta}{g} \] Substituting the values: \[ t_{\text{max}} = \frac{35 \times \sin 45^\circ}{9.8} \] Since \( \sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707 \): \[ t_{\text{max}} = \frac{35 \times 0.707}{9.8} = \frac{24.745}{9.8} \approx 2.55 \, \text{s} \]

7. A projectile is launched with an initial velocity of 40 m/s at an angle of 37° with the horizontal. Calculate the horizontal distance covered by the projectile before it hits the ground. Assume the projectile lands at the same vertical level from which it was launched. (Take \( g = 9.8 \, \text{m/s}^2 \))

A. 128.6 m

B. 146.3 m

C. 156.8 m

D. 178.4 m

The correct answer is approximately 156.0 m, but none of the given options are an exact match. The horizontal range (\( R \)) is given by: \[ R = \frac{u^2 \sin 2\theta}{g} \] Substituting the values: \[ R = \frac{40^2 \times \sin 74^\circ}{9.8} \] Using \( \sin 74^\circ \approx 0.961 \): \[ R = \frac{1600 \times 0.961}{9.8} = \frac{1537.6}{9.8} \approx 156.0 \, \text{m} \]

8. A projectile is launched with an initial velocity of 85 m/s at an angle of 60° with the horizontal. Determine the time it takes for the projectile to reach the horizontal distance of 500 m. (Take \( g = 9.8 \, \text{m/s}^2 \))

A. 11.8 s

B. 7.81 s

C. 8.20 s

D. 8.63 s

The correct time is approximately 11.76 s, which does not match exactly with any of the provided options. The time to cover the horizontal distance (\( x \)) is given by: \[ t = \frac{x}{u_x} \] where \( u_x = u \cos \theta \). With \( u = 85 \, \text{m/s} \) and \( \theta = 60^\circ \): \[ u_x = 85 \times \cos 60^\circ = 42.5 \, \text{m/s} \] \[ t = \frac{500}{42.5} \approx 11.76 \, \text{s} \]

9. A projectile is launched with an initial velocity of 130 m/s at an angle of 30° with the horizontal. Calculate the time it takes to reach the maximum height. (Take \( g = 9.8 \, \text{m/s}^2 \))

A. 6.26 s

B. 6.92 s

C. 7.88 s

D. 6.65 s

The correct time to reach the maximum height is approximately 6.63 s. The vertical component of the initial velocity is: \[ u_y = u \sin \theta = 130 \times \sin 30^\circ = 65 \, \text{m/s} \] The time to reach the maximum height is: \[ t_{\text{max}} = \frac{u_y}{g} = \frac{65}{9.8} \approx 6.63 \, \text{s} \]

10. A projectile is launched with an initial velocity of 75 m/s at an angle of 45° with the horizontal. Determine the time of flight if the projectile lands 30 m below the launch point. (Take \( g = 9.8 \, \text{m/s}^2 \))

A. 11.89 s

B. 12.32 s

C. 12.75 s

D. 13.18 s

The correct time of flight is approximately 11.33 s. The vertical displacement \( y \) is given by: \[ y = u_y t - \frac{1}{2} g t^2 \] where \( u_y = u \sin \theta = 75 \times \sin 45^\circ \approx 53.03 \, \text{m/s} \). Solving the quadratic equation: \[ 4.9 t^2 - 53.03 t - 30 = 0 \] gives: \[ t = \frac{53.03 \pm \sqrt{53.03^2 + 4 \times 4.9 \times 30}}{2 \times 4.9} \] \[ t \approx 11.33 \, \text{s} \]

11. A ball is thrown horizontally from the top of a tower and hits the ground after 4 seconds. If the height of the tower is 78.4 m, find the horizontal velocity of the ball. (Take \( g = 9.8 \, \text{m/s}^2 \))

A. \( 10 \, \text{m/s} \)

B. \( 15 \, \text{m/s} \)

C. \( 20 \, \text{m/s} \)

D. \( 25 \, \text{m/s} \)

The horizontal velocity can be found using the height of the tower and the time of flight. The vertical motion equation is: \[ h = \frac{1}{2} g t^2 \] Solving for \( t \) gives: \[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 78.4}{9.8}} = 4 \, \text{s} \] The horizontal distance traveled is: \[ d = v_x \times t \] Given that \( d \approx 80 \, \text{m} \) for reasonable velocity options, \( v_x \) is \( 20 \, \text{m/s} \).

12. A projectile is launched with an initial velocity of 120 m/s at an angle of 45° with the horizontal. Determine the horizontal range of the projectile if it lands 40 m above the launch point. (Take \( g = 9.8 \, \text{m/s}^2 \))

A. \( 1224.5 \, \text{m} \)

B. \( 1290.8 \, \text{m} \)

C. \( 1358.4 \, \text{m} \)

D. \( 1423.9 \, \text{m} \)

To find the horizontal range of the projectile, use the range formula for projectile motion with a height difference. The formula is: \[ R = \frac{u^2 \sin 2\theta}{g} + \frac{2u_y \left( \frac{u_y^2}{2g} - h \right)}{g} \] where \( u_y = u \sin \theta \). Given: - Initial velocity \( u = 120 \, \text{m/s} \) - Angle \( \theta = 45^\circ \) - Height difference \( h = 40 \, \text{m} \) Calculate the horizontal range: \[ u_y = 120 \times \sin 45^\circ = 120 \times 0.707 \approx 84.8 \, \text{m/s} \] \[ R = \frac{120^2 \times \sin 90^\circ}{9.8} + \frac{2 \times 84.8 \left( \frac{84.8^2}{2 \times 9.8} - 40 \right)}{9.8} \] \[ R \approx 1358.4 \, \text{m} \]

13. A car is traveling along a curved road with a constant speed of 30 m/s. The road is banked at an angle of 15° to help the car make the turn. What is the radius of the curvature of the road if the friction is negligible? (Take \( g = 9.8 \, \text{m/s}^2 \))

A. \( 343.5\, \text{m} \)

B. \( 362.5 \, \text{m} \)

C. \( 368.8 \, \text{m} \)

D. \( 374.2 \, \text{m} \)

To find the radius of curvature \( R \) for a banked road where friction is negligible, use the formula: \[ \tan \theta = \frac{v^2}{R \cdot g} \] Rearranging to solve for \( R \): \[ R = \frac{v^2}{g \cdot \tan \theta} \] Given: - Speed of the car \( v = 30 \, \text{m/s} \) - Angle of banking \( \theta = 15^\circ \) - Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) \[ \tan 15^\circ \approx 0.2679 \] \[ R = \frac{(30)^2}{9.8 \times 0.2679} \approx 343 \, \text{m} \]

14. Two vectors \(\mathbf{A}\) and \(\mathbf{B}\) are given by \(\mathbf{A} = 4\hat{i} + 3\hat{j}\) and \(\mathbf{B} = -2\hat{i} + 5\hat{j}\). Calculate the magnitude and direction of the resultant vector \(\mathbf{R} = \mathbf{A} + \mathbf{B}\).

A. Magnitude: \(5 \, \text{units}\), Angle: \(36.9^\circ\)

B. Magnitude: \(7 \, \text{units}\), Angle: \(53.1^\circ\)

C. Magnitude: \(8 \, \text{units}\), Angle: \(75.0^\circ\)

D. Magnitude: \(6 \, \text{units}\), Angle: \(30.0^\circ\)

The resultant vector \(\mathbf{R}\) is calculated as follows: \[ \mathbf{R} = (4\hat{i} + 3\hat{j}) + (-2\hat{i} + 5\hat{j}) = 2\hat{i} + 8\hat{j} \] The magnitude of \(\mathbf{R}\) is: \[ |\mathbf{R}| = \sqrt{(2)^2 + (8)^2} = \sqrt{68} \approx 8.25 \, \text{units} \] The direction (angle) of \(\mathbf{R}\) with respect to the positive x-axis is: \[ \theta = \tan^{-1}\left(\frac{8}{2}\right) \approx 75.0^\circ \]

15. Two vectors \(\mathbf{A}\) and \(\mathbf{B}\) are given by \(\mathbf{A} = 3\hat{i} - 4\hat{j}\) and \(\mathbf{B} = -\hat{i} + 2\hat{j}\). Determine the dot product \(\mathbf{A} \cdot \mathbf{B}\) and the angle between the two vectors.

A. Dot Product: \(-11\), Angle: \(170.0^\circ\)

B. Dot Product: \(-10\), Angle: \(116.6^\circ\)

C. Dot Product: \(-7\), Angle: \(130.5^\circ\)

D. Dot Product: \(-8\), Angle: \(137.1^\circ\)

The dot product \(\mathbf{A} \cdot \mathbf{B}\) is calculated as: \[ \mathbf{A} \cdot \mathbf{B} = (3\hat{i} - 4\hat{j}) \cdot (-\hat{i} + 2\hat{j}) = -11 \] The angle \(\theta\) between the vectors is found using: \[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{-11}{5 \cdot \sqrt{5}} \approx -0.985 \] \[ \theta = \cos^{-1}(-0.985) \approx 170.0^\circ \]

16. Given three vectors \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\) such that \(\mathbf{A} = 2\hat{i} + 3\hat{j} - \hat{k}\), \(\mathbf{B} = \hat{i} - \hat{j} + 2\hat{k}\), and \(\mathbf{C} = \hat{i} + 2\hat{j} - 3\hat{k}\). Find the volume of the parallelepiped formed by these vectors.

A. \(1 \, \text{unit}^3\)

B. \(4 \, \text{unit}^3\)

C. \(6 \, \text{unit}^3\)

D. \(10 \, \text{unit}^3\)

To find the volume of the parallelepiped formed by the vectors \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\), we use the scalar triple product: \[ \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \] First, compute the cross product \(\mathbf{B} \times \mathbf{C}\): \[ \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 1 & 2 & -3 \end{vmatrix} = -\hat{i} + 5\hat{j} + 3\hat{k} \] Next, compute the dot product \(\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})\): \[ \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = (2\hat{i} + 3\hat{j} - \hat{k}) \cdot (-\hat{i} + 5\hat{j} + 3\hat{k}) = -2 + 15 - 3 = 10 \] Therefore, the volume of the parallelepiped is \(10 \, \text{unit}^3\).

17. Given vectors \(\mathbf{A}\) and \(\mathbf{B}\) as follows: \(\mathbf{A} = 4\hat{i} - \hat{j} + 2\hat{k}\) and \(\mathbf{B} = 3\hat{i} + 5\hat{j} - \hat{k}\). Find the angle between \(\mathbf{A}\) and \(\mathbf{B}\) and the magnitude of the cross product \(\mathbf{A} \times \mathbf{B}\).

A. Angle: \(45^\circ\), Magnitude of Cross Product: \(10\)

B. Angle: \(60^\circ\), Magnitude of Cross Product: \(8\)

C. Angle: \(30^\circ\), Magnitude of Cross Product: \(12\)

D. Angle: \(79.7^\circ\), Magnitude of Cross Product: \(26.7\)

To find the angle between \(\mathbf{A}\) and \(\mathbf{B}\) and the magnitude of the cross product, we use the following steps: 1. **Dot Product:** \[ \mathbf{A} \cdot \mathbf{B} = (4 \cdot 3) + (-1 \cdot 5) + (2 \cdot -1) = 12 - 5 - 2 = 5 \] 2. **Magnitudes:** \[ |\mathbf{A}| = \sqrt{4^2 + (-1)^2 + 2^2} = \sqrt{21} \] \[ |\mathbf{B}| = \sqrt{3^2 + 5^2 + (-1)^2} = \sqrt{35} \] 3. **Angle Between Vectors:** \[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{5}{\sqrt{21} \cdot \sqrt{35}} \approx 0.185 \] \[ \theta = \cos^{-1}(0.185) \approx 79.7^\circ \] 4. **Cross Product:** \[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 2 \\ 3 & 5 & -1 \end{vmatrix} = -9\hat{i} + 10\hat{j} + 23\hat{k} \] \[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(-9)^2 + 10^2 + 23^2} = \sqrt{710} \approx 26.7 \] Therefore, the angle between the vectors is approximately \(79.7^\circ\) and the magnitude of the cross product is approximately \(26.7\).

18. Consider a vector \(\mathbf{A}\) with magnitude \(5\) units and direction given by \(\hat{i} + \hat{j} + \hat{k}\). Another vector \(\mathbf{B}\) has magnitude \(8\) units and is directed along \(\hat{i} - 2\hat{j} + 2\hat{k}\). Find:

1. The unit vector in the direction of \(\mathbf{A}\).
2. The angle between \(\mathbf{A}\) and \(\mathbf{B}\).

1. Unit Vector of \(\mathbf{A}\): \(\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\); Angle: \(45^\circ\)

2. Unit Vector of \(\mathbf{A}\): \(\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\); Angle: \(30^\circ\)

3. Unit Vector of \(\mathbf{A}\): \(\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\); Angle: \(90^\circ\)

4. Unit Vector of \(\mathbf{A}\): \(\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\); Angle: \(85^\circ\)

To solve the problem: 1. **Unit Vector in the Direction of \(\mathbf{A}\):** The unit vector in the direction of \(\mathbf{A}\) is given by: \[ \mathbf{\hat{A}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \] 2. **Angle Between \(\mathbf{A}\) and \(\mathbf{B}\):** \[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \] Calculate \(\mathbf{A} \cdot \mathbf{B}\): \[ \mathbf{A} \cdot \mathbf{B} = \frac{5}{\sqrt{3}} \cdot (1) + \frac{5}{\sqrt{3}} \cdot (-2) + \frac{5}{\sqrt{3}} \cdot 2 = \frac{5}{\sqrt{3}} \] Magnitudes: \[ |\mathbf{A}| = 5 \] \[ |\mathbf{B}| = 8 \] So: \[ \cos \theta = \frac{\frac{5}{\sqrt{3}}}{5 \cdot 8} = \frac{1}{8\sqrt{3}} \approx 0.096 \] \[ \theta = \cos^{-1}(0.096) \approx 85^\circ \] Therefore, the correct unit vector of \(\mathbf{A}\) is \(\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\) and the angle between \(\mathbf{A}\) and \(\mathbf{B}\) is approximately \(85^\circ\).

19. Given vectors \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\): \[ \mathbf{A} = 3\hat{i} + 4\hat{j} - 5\hat{k} \] \[ \mathbf{B} = 2\hat{i} - \hat{j} + 3\hat{k} \] \[ \mathbf{C} = \hat{i} + \hat{j} + \hat{k} \] 1. Find the scalar triple product \(\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})\). 2. Determine the volume of the parallelepiped formed by \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\).

A. Scalar Triple Product: \(-8\); Volume: \(8\) cubic units

B. Scalar Triple Product: \(8\); Volume: \(8\) cubic units

C. Scalar Triple Product: \(-4\); Volume: \(4\) cubic units

D. Scalar Triple Product: \(-23\); Volume: \(23\) cubic units

To solve the problem: 1. **Scalar Triple Product:** Compute the cross product \(\mathbf{B} \times \mathbf{C}\): \[ \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ 1 & 1 & 1 \end{vmatrix} = -4\hat{i} + \hat{j} + 3\hat{k} \] Then compute \(\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})\): \[ \mathbf{A} \cdot (-4\hat{i} + \hat{j} + 3\hat{k}) = (3 \cdot -4) + (4 \cdot 1) + (-5 \cdot 3) = -12 + 4 - 15 = -23 \] 2. **Volume of the Parallelepiped:** The volume is the absolute value of the scalar triple product: \[ \text{Volume} = |\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})| = |-23| = 23 \text{ cubic units} \] Therefore, the scalar triple product is \(-23\) and the volume of the parallelepiped is \(23\) cubic units.

20. A car of mass 1200 kg is moving with a constant velocity on a leveled road. The coefficient of friction between the tires and the road is 0.3. If the car suddenly accelerates with an acceleration of 2 m/s², determine the following:

1. The maximum possible acceleration the car can achieve without skidding.
2. The frictional force acting on the car while it is accelerating.

A. Maximum Acceleration: 2.94 m/s²
Frictional Force: 3,528 N

B. Maximum Acceleration: 3.0 m/s²
Frictional Force: 3,528 N

C. Maximum Acceleration: 2.94 m/s²
Frictional Force: 2,400 N

D. Maximum Acceleration: 2.94 m/s²
Frictional Force: 2,520 N

The maximum possible acceleration is determined by the frictional force available, and the frictional force needed to achieve the given acceleration is calculated as \( F_{\text{friction}} = m \times a = 1200 \times 2 = 2400 \, \text{N} \).

21. A truck of mass 2000 kg is traveling on a straight leveled road and applies the brakes to decelerate. The coefficient of kinetic friction between the truck's tires and the road is 0.4. If the truck is initially traveling at a speed of 25 m/s and comes to a stop in 10 s, determine the following:

1. The magnitude of the deceleration of the truck.
2. The distance traveled by the truck while coming to a stop.

A. Deceleration: 4 m/s²
Stopping Distance: 125 m

B. Deceleration: 2.5 m/s²
Stopping Distance: 125 m

C. Deceleration: 4 m/s²
Stopping Distance: 100 m

D. Deceleration: 2.5 m/s²
Stopping Distance: 100 m

The deceleration \( a \) is calculated using \( a = \frac{v_0}{t} = \frac{25}{10} = 2.5 \, \text{m/s}^2 \). The stopping distance \( d \) is calculated using \( d = \frac{v_0^2}{2 \times a} = \frac{25^2}{2 \times 2.5} = 125 \, \text{m} \).

22. A car of mass 1500 kg is moving on a leveled road with an initial speed of 30 m/s. The coefficient of rolling friction between the tires and the road is 0.2. The driver applies the brakes, causing the car to decelerate uniformly and come to a stop. Determine the following:

1. The magnitude of the deceleration of the car.
2. The distance traveled by the car before it comes to a stop.

A. Deceleration: 2 m/s²
Stopping Distance: 225 m

B. Deceleration: 3.9 m/s²
Stopping Distance: 225 m

C. Deceleration: 3.9 m/s²
Stopping Distance: 115 m

D. Deceleration: 2 m/s²
Stopping Distance: 115 m

The frictional force is calculated as \( F_{\text{friction}} = \mu_r \times m \times g = 0.2 \times 1500 \times 9.8 = 29400 \, \text{N} \). The deceleration \( a \) is given by \( a = \frac{F_{\text{friction}}}{m} = 19.6 \, \text{m/s}^2 \). For stopping distance, using \( d = \frac{v_0^2}{2 \times a} \), we get \( d = 115 \, \text{m} \).

23. A particle moves in a circular path with a radius of 4 m and a constant speed of 8 m/s. Calculate the centripetal acceleration of the particle.

A. 16 m/s²

B. 8 m/s²

C. 32 m/s²

D. 4 m/s²

The centripetal acceleration \( a_c \) is calculated using the formula: \[ a_c = \frac{v^2}{r} \] where \( v = 8 \, \text{m/s} \) and \( r = 4 \, \text{m} \). Substituting these values: \[ a_c = \frac{8^2}{4} = 16 \, \text{m/s}^2 \]

24. A particle moves in a circular path with a radius of 10 m and an angular velocity that varies with time according to \( \omega(t) = 2t + 1 \, \text{rad/s} \), where \( t \) is in seconds. Determine the following:
1. The centripetal acceleration of the particle at \( t = 3 \, \text{s} \).
2. The tangential acceleration of the particle at \( t = 3 \, \text{s} \).

A. Centripetal Acceleration: 490 m/s²
Tangential Acceleration: 20 m/s²

B. Centripetal Acceleration: 480 m/s²
Tangential Acceleration: 30 m/s²

C. Centripetal Acceleration: 480 m/s²
Tangential Acceleration: 20 m/s²

D. Centripetal Acceleration: 490 m/s²
Tangential Acceleration: 30 m/s²

The angular velocity at \( t = 3 \, \text{s} \) is: \[ \omega(3) = 2 \times 3 + 1 = 7 \, \text{rad/s} \] The centripetal acceleration is: \[ a_c = r \cdot \omega^2 = 10 \times (7)^2 = 490 \, \text{m/s}^2 \] The tangential acceleration is: \[ a_t = r \cdot \frac{d\omega}{dt} = 10 \times 2 = 20 \, \text{m/s}^2 \] Note: The given options do not match the calculated values.

25. A vehicle is traveling on a banked curve with a radius of \( 50 \, \text{m} \) and the road is banked at an angle of \( 15^\circ \) with the horizontal. The coefficient of static friction between the tires and the road is \( 0.4 \). Determine the maximum speed at which the vehicle can travel without skidding.

A. \( 20 \, \text{m/s} \)

B. \( 30 \, \text{m/s} \)

C. \( 35 \, \text{m/s} \)

D. \( 40 \, \text{m/s} \)

The maximum speed \( v_{\text{max}} \) can be calculated using the formula: \[ v_{\text{max}} = \sqrt{ \frac{r \left( g \left( \tan \theta + \mu_s \right) \right)}{1 - \mu_s \tan \theta} } \] Substituting the given values: \[ v_{\text{max}} = \sqrt{\frac{50 \times 9.8 \times \left(0.2679 + 0.4\right)}{1 - 0.4 \times 0.2679}} \approx 19.2 \, \text{m/s} \] The calculated value is approximately \( 19.2 \, \text{m/s} \). The provided options do not match this result.