**1. A cycle is travelling along a straight road. It accelerates from 2 m/s to 12 m/s in 5 seconds. At the same time, a car starts from rest and accelerates uniformly at 20 m/s². After how much time will the car catch up to the cyclist?**

a) 5.25 s

b) 5 s

c) 10 s

d) 2.45 s

**Answer:** b) To solve this problem, let's first calculate the distance covered by both the cycle and the car during the given time period, and then find out when they meet.

Given:

Initial velocity of the cyclist (u_{1}) = 2 ms^{-1}

Final velocity of the cyclist (v_{1}) = 12 ms^{-1}

Time (t_{1})= 5 seconds

Car's initial velocity (u_{2}) = 0 ms^{-1} (starts from rest)

Car's acceleration (a_{2}) = 20 ms^{-2}

We want to find time t when the car catches up to the cyclist.

For the cyclist:

Using the following equation we can calculate the acceleration (a_{1}) of the cyclist:

a_{1} = (v_{1} - u_{1})/t_{1}

a_{1} = (12 ms^{-1} - 2 ms^{-1})/ 5 s

a_{1} = 2 ms^{-2}

Using the second equation of motion we can calculate the distance travelled by the cyclist (s_{1}):

s_{1} = u_{1} x t + 1/2 a_{1} x t^{2}

s_{1} = 2 x t + 1/2 x 2 x t^{2}

s_{1} = 2t x t^{2}

For the car:

Using the second equation of motion we can find the distance travelled by car (s_{2}):

s_{2} = u_{2} x t + 1/2 a_{2} x t^{2}

s_{2} = 0 + 1/2 x 20 x t^{2}

s_{2} = 10t^{2}

Now, we can find the time t when the distances are equal:

s_{2} = s_{1}

10t^{2} = 2t x t^{2}

t = 5 s

The car will catch up to the cyclist after 5 seconds.

**2. How does the velocity-time graph of an object in free fall (with constant acceleration due to gravity) appear?**

a)

b)

c)

d)

**Answer:** d) When an object is in free fall and experiencing constant acceleration due to gravity, its velocity changes uniformly over time. As the object falls, its velocity increases in the downward direction. This results in a straight line on a velocity-time graph with a negative slope as depicted by option d. The steeper the slope, the greater the acceleration due to gravity.

**3. A train decelerates at a rate of 2 ms ^{-2} and comes to a stop. If its initial velocity was 30 ms^{-1}, how far does it travel during the deceleration?**

a) 345 m

b) -225 m

c) 225 m

d) -60 m

**Answer:** c) To solve this problem, we can use the equation of motion that relates initial velocity, final velocity, acceleration, and distance:

Given:

Initial velocity (u) = 30 ms^{-1 }(positive because it's in the forward direction)

Final velocity (v) = 0 ms^{-1} (the train comes to a stop)

Acceleration (a) = -2 ms^{-2} (negative because it's deceleration)

We need to find the distance (s) travelled during the deceleration.

We'll use the equation of motion:

v^{2} = u^{2} + 2as

Substitute the given values:

(0)^{2} = (30^{2}) + 2 x (−2) x s

0 = 900 − 4s

Solving for s:

4s = 900

s = 900/4

s = 225 m

So, the train travels a distance of 225 m during the deceleration.

**4. A person is sitting in a train that's moving at a constant velocity. They toss a ball vertically upward. How does the ball's motion look to an observer outside the moving train?**

a) The ball follows a curved path due to the train's motion.

b) The ball follows a straight vertical path upward and downward.

c) The ball follows a parabolic path upward and downward.

d) The ball's motion is unaffected by the train's velocity.

**Answer:** b) When the train is moving at a constant velocity, the person inside the train, as well as the ball they toss vertically upward, are also moving with that constant velocity. From the perspective of an observer outside the train, the ball's vertical motion is independent of the train's horizontal motion.

Gravity acts vertically downward on the ball, causing it to follow a straight vertical path upward and then downward. This is because the ball's motion is determined by the force of gravity and its initial upward velocity. The train's constant velocity does not influence the vertical motion of the ball, and it will appear to move vertically straight to an observer outside the train.

**5. A Ferris wheel has a radius of 15 m and completes one revolution in 60 seconds. Calculate the speed of a person sitting on the Ferris wheel at the outer edge.**

a) 0.5 ms^{-1}

b) 1.5 ms^{-1}

c) 1 ms^{-1}

d) 2 ms^{-1}

**Answer:** b) To calculate the linear speed of a person sitting on the Ferris wheel at the outer edge, we can use the formula for calculating the velocity of an object undergoing uniform circular motion:

v = Circumference/Time taken

Given:

Radius of the Ferris wheel (r) = 15 metres

Time taken for one revolution (t) = 60 seconds

Substitute the given values into the formula:

v = (2π×15)/60

v = 30π/60

v = 0.5π

Since the value of π is approximately 3.14, we can approximate speed:

v ≈ 0.5 × 3.14

v ≈ 1.57 ms-1

So, the speed of a person sitting on the Ferris wheel at the outer edge is approximately 1.57 ms^{-1}.

**1. What is Velocity and How Does it Differ from Speed?**

Velocity is defined as the rate at which an object's displacement changes over time. Velocity, unlike speed, incorporates both the distance traveled and the direction of motion. It is a vector quantity, therefore it has both magnitude (speed) and direction.

**2. What Factors Affect the Motion of Objects?**

Several factors might influence object motion, including force, mass, friction, gravity, and air resistance. Forces can force an item to accelerate, decelerate, change direction, or remain stationary. Understanding these elements aids in explaining and predicting the behaviour of moving objects in a variety of scenarios.

**3. How Can Graphs Be Used to Represent Motion?**

Distance-time graphs and velocity-time graphs are typical representations of object motion. Distance-time graphs illustrate how the distance traveled by an item varies over time, whereas velocity-time graphs show how an object's velocity changes over time. Analyzing these graphs allows us to visualize and analyze the motion of things.

**4. How Do Newton's Laws of Motion Explain the Behavior of Objects?**

Newton's Laws of Motion are fundamental concepts that define how things move and the forces that act on them. These laws include Newton's First Law (the law of inertia), Second Law (the law of acceleration), and Third Law (the action-reaction principle). They give a framework for comprehending the causes and implications of motion in everyday settings.

**5. What Are the Different Types of Motion?**

Motion is classified into four types: **linear motion** (motion along a straight route), **rotational motion** (motion around an axis), **oscillatory motion** (repetitive back-and-forth motion), and **projectile motion** (motion of an item propelled into the air under the effect of gravity). Each form of motion possesses distinct properties and behaviors.

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