krit.club logo

Physics - Motion, forces, and energy (Velocity, Acceleration, and Newton's Laws)

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Speed is a scalar quantity representing the distance traveled per unit time: v=dtv = \frac{d}{t}.

Velocity is a vector quantity that describes speed in a given direction. A change in direction results in a change in velocity even if speed is constant.

Acceleration (aa) is defined as the rate of change of velocity: a=Δvta = \frac{\Delta v}{t}. It is measured in units of m/s2\text{m/s}^2.

In a Distance-Time graph, the gradient (slope) represents the speed of the object.

In a Speed-Time graph, the gradient represents the acceleration, and the area under the curve represents the total distance traveled.

Newton's First Law (Inertia): An object remains at rest or continues at a constant velocity unless acted upon by a resultant (net) force.

Newton's Second Law: The acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass: F=maF = ma.

Newton's Third Law: For every action force, there is an equal and opposite reaction force acting on a different body.

Mass is the amount of matter in an object (measured in kg\text{kg}), while Weight is the force of gravity acting on that mass: W=mgW = mg (where g9.8 m/s2g \approx 9.8\text{ m/s}^2 on Earth).

📐Formulae

v=stv = \frac{s}{t}

a=vuta = \frac{v - u}{t}

F=m×aF = m \times a

W=m×gW = m \times g

p=m×vp = m \times v

💡Examples

Problem 1:

A cyclist accelerates from a velocity of 4 m/s4\text{ m/s} to 12 m/s12\text{ m/s} in a time of 5 seconds5\text{ seconds}. Calculate the acceleration.

Solution:

a=vut=12 m/s4 m/s5 s=85=1.6 m/s2a = \frac{v - u}{t} = \frac{12\text{ m/s} - 4\text{ m/s}}{5\text{ s}} = \frac{8}{5} = 1.6\text{ m/s}^2

Explanation:

To find acceleration, subtract the initial velocity (uu) from the final velocity (vv) and divide by the time taken (tt).

Problem 2:

A car of mass 1200 kg1200\text{ kg} is subjected to a constant resultant force of 3600 N3600\text{ N}. Determine the acceleration of the car.

Solution:

a=Fm=3600 N1200 kg=3 m/s2a = \frac{F}{m} = \frac{3600\text{ N}}{1200\text{ kg}} = 3\text{ m/s}^2

Explanation:

Using Newton's Second Law (F=maF = ma), we rearrange the formula to solve for acceleration by dividing the force by the mass.

Problem 3:

Calculate the weight of an astronaut with a mass of 80 kg80\text{ kg} on the Moon, where the gravitational field strength is gmoon=1.6 m/s2g_{moon} = 1.6\text{ m/s}^2.

Solution:

W=m×g=80 kg×1.6 m/s2=128 NW = m \times g = 80\text{ kg} \times 1.6\text{ m/s}^2 = 128\text{ N}

Explanation:

Weight is a force calculated by multiplying the mass of the object by the local gravitational field strength (gg).