Motion in a Plane Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For an object in motion with constant acceleration in a plane, what is the shape of its path if the initial velocity is zero?

If the initial velocity is zero, the object moves under constant acceleration alone, resulting in a straight-line path along the direction of the acceleration vector. This is a special case of rectilinear motion.

Parabola
Circle
Straight line
Ellipse
3

A ball is thrown at \( 24 \, \text{m/s} \) at \( 37^\circ \). What is its vertical velocity at \( t = 1 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2, \sin 37^\circ = 0.6 \))

Vertical velocity \( v_y = v_0 \sin \theta_0 - g t \).

Given: \( v_0 = 24 \, \text{m/s}, \sin 37^\circ = 0.6, t = 1 \, \text{s}, g = 10 \, \text{m/s}^2 \).

\( v_y = 24 \times 0.6 - 10 \times 1 = 14.4 - 10 = 4.4 \, \text{m/s} \).

5 m/s
10 m/s
4.4 m/s
0 m/s
3

A particle moves in a circular path of radius \( 1 \, \text{m} \) completing 3 revolutions in \( 6 \, \text{s} \). What is its centripetal acceleration?

Frequency \( f = \frac{n}{t} = \frac{3}{6} = 0.5 \, \text{Hz} \).

Angular speed \( \omega = 2\pi f = 2\pi \times 0.5 = \pi \, \text{rad/s} \).

Centripetal acceleration \( a_c = \omega^2 R = (\pi)^2 \times 1 \approx 9.87 \, \text{m/s}^2 \).

9.87 m/s²
5 m/s²
15 m/s²
20 m/s²
1

For an object in motion with constant acceleration in a plane, what is true when the acceleration is parallel to the initial velocity?

When the constant acceleration is parallel to the initial velocity, the motion occurs in a straight line along the direction of the initial velocity, as the acceleration only changes the magnitude of the velocity, not its direction.

The path is a straight line
The path is parabolic
The path is circular
The path is elliptical
1

A projectile launched at \( 15 \, \text{m/s} \) at \( 53^\circ \) reaches a horizontal distance of \( 18 \, \text{m} \). What is the value of \( g \)? (Take \( \sin 53^\circ = 0.8, \cos 53^\circ = 0.6 \))

Range \( R = \frac{v_0^2 \sin 2\theta_0}{g} \), where \( 2\theta_0 = 106^\circ, \sin 106^\circ = \sin (180^\circ - 106^\circ) = \sin 74^\circ \approx 0.96 \).

Given: \( R = 18 \, \text{m}, v_0 = 15 \, \text{m/s} \).

\( 18 = \frac{15^2 \times 0.96}{g} = \frac{225 \times 0.96}{g} = \frac{216}{g} \).

\( g = \frac{216}{18} = 12 \, \text{m/s}^2 \).

12 m/s²
10 m/s²
9.8 m/s²
15 m/s²
1

A particle moves in a circle of radius \( 2.5 \, \text{m} \) with a speed of \( 10 \, \text{m/s} \). What is its centripetal acceleration?

Centripetal acceleration \( a_c = \frac{v^2}{R} \).

Given: \( v = 10 \, \text{m/s}, R = 2.5 \, \text{m} \).

\( a_c = \frac{10^2}{2.5} = \frac{100}{2.5} = 40 \, \text{m/s}^2 \).

20 m/s²
30 m/s²
50 m/s²
40 m/s²
4

A particle’s position is given by \( x = 4t^2 - t \) and \( y = 2t + t^2 \) (in meters and seconds). What is its speed at \( t = 1 \, \text{s} \)?

Velocity: \( v_x = \frac{dx}{dt} = 8t - 1, v_y = \frac{dy}{dt} = 2 + 2t \).

At \( t = 1 \, \text{s} \): \( v_x = 8 \times 1 - 1 = 7 \, \text{m/s}, v_y = 2 + 2 \times 1 = 4 \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{7^2 + 4^2} = \sqrt{49 + 16} = \sqrt{65} \approx 8.06 \, \text{m/s} \).

6 m/s
8.06 m/s
10 m/s
7 m/s
2

For a projectile launched from the ground, what is true about its time of ascent and time of descent?

For a projectile launched from and returning to the ground, the time of ascent equals the time of descent due to the symmetry of the parabolic path. The time to reach maximum height (\( t_m = \frac{v_0 \sin \theta_0}{g} \)) is half the total time of flight, and the descent mirrors the ascent.

Ascent time is greater
Descent time is greater
They are equal
Depends on the launch angle
3

A projectile is launched at \( 16 \, \text{m/s} \) at \( 37^\circ \). What is its horizontal velocity component? (Take \( \cos 37^\circ = 0.8 \))

Horizontal velocity \( v_x = v_0 \cos \theta_0 \).

Given: \( v_0 = 16 \, \text{m/s}, \cos 37^\circ = 0.8 \).

\( v_x = 16 \times 0.8 = 12.8 \, \text{m/s} \).

10 m/s
16 m/s
12.8 m/s
8 m/s
3

A particle starts with velocity \( 6 \hat{i} \, \text{m/s} \) and accelerates at \( (2 \hat{i} + 4 \hat{j}) \, \text{m/s}^2 \). What is its speed after \( 3 \, \text{s} \)?

Velocity \( \mathbf{v} = \mathbf{v}_0 + \mathbf{a} t \).

Given: \( \mathbf{v}_0 = 6 \hat{i}, \mathbf{a} = 2 \hat{i} + 4 \hat{j}, t = 3 \, \text{s} \).

\( \mathbf{v} = 6 \hat{i} + (2 \hat{i} + 4 \hat{j}) \times 3 = 6 \hat{i} + 6 \hat{i} + 12 \hat{j} = 12 \hat{i} + 12 \hat{j} \, \text{m/s} \).

Speed \( v = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \approx 16.97 \, \text{m/s} \).

12 m/s
15 m/s
16.97 m/s
20 m/s
3

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