Rectilinear Motion

Question 1

Differentiate distance from displacement.

Answer 1

Distance is a scalar quantity that measures the total path traveled by the particle.

Displacement is a vector quantity that represents the change in position of a particle from its initial to its final point.

Displacement considers direction, measuring the shortest path, while distance is the total length traveled, ignoring direction.

Question 2

Define:

velocity

Answer 2

A vector quantity that describes the rate of change of displacement with respect to time, including both speed and direction.

Average velocity is total displacement divided by total time, giving an overall measure of motion. Instantaneous velocity is the velocity at a specific moment, found as the derivative of displacement with respect to time.

Question 3

Define:

acceleration

Answer 3

The rate of change of velocity with respect to time, indicating how quickly an object’s speed or direction changes.

Average acceleration is the change in velocity over the total time taken, while instantaneous acceleration is the rate of change of velocity at a specific moment in time, often calculated as the derivative of velocity with respect to time.

Question 4

What is the formula for final position in rectilinear motion with constant acceleration, considering the initial velocity, time, and acceleration?

Answer 4

𝑥 = 𝑥₀ + 𝑣₀𝑡 + ½𝑎𝑡²

where:
𝑥 = final position
𝑥₀ = initial position
𝑣₀ = initial velocity
𝑡 = time
𝑎 = acceleration

Question 5

What is the formula for velocity in uniformly accelerated motion, derived from the definition of acceleration?

Answer 5

𝑣 = 𝑣₀ + 𝑎 ⋅ 𝑡

The definition of acceleration:
𝑎 = 𝑑𝑣 / 𝑑𝑡

Rearrange to express velocity change:
𝑑𝑣 = 𝑎 𝑑𝑡

Integrate both sides:
∫ 𝑑𝑣 = 𝑎 ∫ 𝑑𝑡

The acceleration a is constant, so the integral becomes:
𝑣 − 𝑣₀ = 𝑎 ⋅ 𝑡

Thus, the velocity equation is:
𝑣 = 𝑣₀ + 𝑎 ⋅ 𝑡

Question 6

Solve the problem:

An object starts from rest and accelerates uniformly at 3 m/s². What will be its velocity after 5 seconds?

Answer 6

15 m/s

Using the equation for velocity:
𝑣 = 𝑣₀ + 𝑎𝑡

Since the object starts from rest, 𝑣₀ = 0, so:
𝑣 = 0 + (3)(5) = 15 m/s

Question 7

Solve the problem:

A car travels a distance of 100 meters in 20 seconds. What is its average velocity?

Answer 7

5 m/s

The formula for average velocity:
𝑣ₐᵥ𝑔 = Δ𝑥 / Δ𝑡

𝑣ₐᵥ𝑔 = 100 m / 20 s = 5 m/s

Question 8

Solve the problem:

An object is moving with an initial velocity of 10 m/s and accelerates at 2 m/s² for 4 seconds. What will be the displacement?

Answer 8

56 m

Using the displacement equation:
𝑥 = 𝑥₀ + 𝑣₀𝑡 + ½𝑎𝑡²

Assume 𝑥₀ = 0, so:
𝑥 = 0 + (10)(4) + ½(2)(4)² = 40 + 16 = 56 m

Question 9

Solve the problem:

A ball is thrown vertically upward with an initial velocity of 20 m/s. What will be the maximum height it reaches?

Assume acceleration due to gravity, 𝑔 = 9.8 m/s².

Answer 9

20.4 m

At maximum height, the final velocity 𝑣 = 0. Using the equation for velocity:
𝑣 = 𝑣₀ − 𝑔𝑡

Set 𝑣 = 0:
0 = 20 − 9.8𝑡
𝑡 = 20 / 9.8 ≈ 2.04 seconds

Now, using the displacement equation:
𝑥 = 𝑥₀ + 𝑣₀𝑡 − ½𝑔𝑡²
Assume 𝑥₀ = 0:
𝑥 = 0 + (20)(2.04) − ½(9.8)(2.04)² ≈ 40.8 − 20.4 = 20.4 m

Question 10

Solve the problem:

An object is moving with a velocity of 30 m/s and decelerates at 2 m/s². How long will it take to stop?

Answer 10

15 seconds

Using the equation for velocity:
𝑣 = 𝑣₀ + 𝑎𝑡

Set 𝑣 = 0 (since the object stops):
0 = 30 + (−2)𝑡
𝑡 = 30 / 2 = 15 seconds