Chapter 3 #1

Question 1

Define Displacement.

Answer 1

Distance from a point in a straight line in a given direction (vector).

Question 2

Define Distance

Answer 2

The total length of the path moved by an object. It is a scalar quantity.

Question 3

Define Speed

Answer 3

The distance traveled per unit time. It is a scalar quantity.

Question 4

Define Velocity

Answer 4

The rate of change of displacement. It is a vector quantity.

Question 5

Define Acceleration

Answer 5

The rate of change of velocity. It is a vector quantity.

Question 6

Formula for average speed under constant acceleration?

Answer 6

(u + v) ÷ 2.

Question 7

Gradient of displacement-time graph?

Answer 7

Velocity.

Question 8

Decreasing gradient on displacement-time graph?

Answer 8

Deceleration.

Question 9

Gradient and area under velocity-time graph?

Answer 9

Gradient: acceleration; Area: displacement.

Question 10

Horizontal line on velocity-time graph?

Answer 10

Constant velocity.

Question 11

Area under acceleration-time graph?

Answer 11

Change in velocity.

Question 12

List the 4 suvat equations.

Answer 12

1. v = u + at
2. v^2 = u^2 + 2as
3. s = ut + ½at^2
4. s = (u + v)/2 × t.

Question 13

What are the two independent components of motion in projectiles?

Answer 13

1. Horizontal Motion: Constant velocity (acceleration = 0).
2. Vertical Motion: Constant acceleration (acceleration = g = 9.81 m s⁻²).

Question 14

For Case 1: Horizontal Projection (e.g., kicking a ball off a cliff), what are the initial velocities?

Answer 14

1. Initial Horizontal Velocity (u_x) = u
2. Initial Vertical Velocity (u_y) = 0

Question 15

For Case 1: Horizontal Projection, what are the formulas for displacement at time t?

Answer 15

1. Horizontal distance (x) = u × t
2. Vertical distance (y) = ½g t^2

Question 16

For Case 1: Horizontal Projection, how do you calculate the final vertical velocity (v_y) at time t?

Answer 16

v_y = g × t

Question 17

For Case 2: Projection at an Angle θ to the horizontal, how do you resolve the initial velocity u

Answer 17

1. Horizontal component: u_x = u cos(θ)
2. Vertical component: u_y = u sin(θ)

Question 18

For Case 2: Projection at an Angle θ to the horizontal, how do you resolve the initial velocity u

Answer 18

1. Horizontal component: u_x = u cos(θ)
2. Vertical component: u_y = u sin(θ)

Question 19

For a projectile launched at an angle, what is the formula for the horizontal distance (range)?

Answer 19

Distance = (u cos(θ)) × t

Question 20

How do you calculate the Time to Maximum Height for a projectile launched at an angle?

Answer 20

Use the vertical component (u sin θ) and the equation
v = u + at:
1. 0 = (u sin θ) - gt
2. t = (u sin θ) / g

Question 21

If a projectile lands at the same level it was launched from, what is the Total Time of Flight?

Answer 21

It is 2 times the time taken to reach maximum height.

Question 22

How do you calculate the Maximum Height (S) reached?

Answer 22

Use the vertical component and the equation v^2 = u^2 + 2as:
0 = (u sin θ)^2 - 2gH

Question 23

Acceleration of object thrown up, with no air resistance?

Answer 23

Constant −9.81 m s^-2 downward.

Question 24

In projectile motion, what is the value of the horizontal acceleration (neglecting air resistance)?

Answer 24

0 m s^-2 (The horizontal velocity remains constant).

Question 25

Vertical velocity at maximum height [Projectiles]?

Answer 25

0 m s^-1

Question 26

What is the formula for the Horizontal Range (distance) of a projectile?

Answer 26

s = ut

Question 27

How does air resistance affect the shape of a projectile's trajectory?

Answer 27

The path is no longer a symmetric parabola; the maximum height is lower, the range is shorter, and the angle of descent is steeper than the angle of launch. If dropped from a high enough height, the angle that the object reaches the ground with is 90°, due to terminal velocity.

Question 28

In projectile motion, why does the vertical component of velocity change?

Answer 28

Due to the acceleration of free fall (gravity) acting vertically.

Question 29

Explain terminal velocity.

Answer 29

1. Drag increases 2. Net force decreases 3. Acceleration decreases until drag = weight 4. Terminal velocity reached

Question 30

Displacement from velocity-time graph with bounce?

Answer 30

Areas above axis positive, below negative.

Question 31

Why is projectile path parabolic?

Answer 31

Constant horizontal velocity and constant vertical acceleration.

Question 32

Does launch angle affect final impact speed?

Answer 32

No; energy conservation.

Question 33

How does drag depend on velocity?

Answer 33

Drag ∝ v^2.