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Displacement, distance, speed and velocity IB DP Physics Study Notes - 2025 Syllabus

Displacement, distance, speed and velocity IB DP Physics Study Notes

Displacement, distance, speed and velocity IB DP Physics Study Notes at  IITian Academy  focus on  specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand

  • that the motion of bodies through space and time can be described and analysed in terms of position,
    velocity, and acceleration
  • velocity is the rate of change of position, and acceleration is the rate of change of velocity
  •  the change in position is the displacement
  •  the difference between distance and displacement
  •  the difference between instantaneous and average values of velocity, speed and acceleration, and how to determine them

Standard level and higher level: 9 hours
Additional higher level: There is no additional higher level content

IB DP Physics 2025 -Study Notes -All Topics

Distance and displacement

∙Mechanics is the branch of physics which concerns itself with forces, and how they affect a body’s motion.
∙Kinematics is the sub-branch of mechanics which studies only a body’s motion without regard to causes.
∙Dynamics is the sub-branch of mechanics which studies the forces which cause a body’s motion.

Kinematics is the study of displacement, velocity and acceleration, or in short, a study of motion.
∙A study of motion begins with position and change in position.
∙Consider Freddie the Fly, and his quest for food:

∙The distance Freddie travels is simply how far he has flown, without regard to direction. Freddie’s distance is 6 meters.

∙Distance is simply how far something has traveled without regard to direction. Freddy has gone 6 m.
∙Displacement, on the other hand, is not only distance traveled, but also direction.

∙This makes displacement a vector. It has a magnitude (6 m) and a direction (+ x-direction).
∙We say Freddie travels through a displacement of 6 m in the positive x-direction.

∙Let’s revisit some previous examples of a ball moving through some displacements…

∙Displacement A is just 15 m to the right (or +15 m for short).
∙Displacement B is just 20 m to the left (or -20 m for short).

FYI
∙Distance A is 15 m, and Distance B is 20 m. There is no regard for direction in distance.
∙Now for some detailed analysis of these two motions…

∙Displacement ∆x (or s) has the following formulas:

FYI

∙Many textbooks use ∆x for displacement, and IB uses s. Don’t confuse the “change in ∆” with the “uncertainty ∆” symbol. And don’t confuse s with seconds!

EXAMPLE:

Use the displacement formula to find each displacement. Note that the x = 0 coordinate has been placed on the number lines.

▶️Answer/Explanation

SOLUTION:

For A: s = (+10) – (-5) = +15 m.
For B: s = (-10) – (+10) = -20 m.

FYI

∙The correct direction (sign) is automatic!

Speed and velocity

Velocity $v$ is a measure of how fast an object moves through a displacement.
Thus, velocity is displacement divided by time, and is measured in meters per second $\left(\mathrm{m} \mathrm{s}^{-1}\right)$.

 

EXAMPLE:

Find the velocity of the second ball (Ball B) if it takes 4 seconds to complete its displacement.
SOLUTION:

For B: $s=(-10)-(+10)=-20 \mathrm{~m}$.
But $t=4 \mathrm{~s}$. Therefore $v=-\frac{20 m}{4 \mathrm{~s}}=-5 \mathrm{~m} \mathrm{~s}^{-1}$.

From the previous example we calculated the velocity of the ball to be -5 m /s.

∙Thus, the ball is moving 5 m /s to the left.
∙With disregard to the direction, we can say that the ball’s speed is 5 m/ s.
∙We define speed as distance divided by time, with disregard to direction.

PRACTICE:

A runner travels 64.5 meters in the negative x-direction in 31.75 seconds. Find her velocity, and her speed.

SOLUTION:
• Her velocity is \( -\frac{64.5}{31.75} = -2.03 \) m/s.

• Her speed is \( \frac{64.5}{31.75} = 2.03 \) m/s.

Acceleration

Acceleration is the change in velocity over time.

 Since $u$ and $v$ are measured in $\frac{m}{s}$ and since $t$ is measured in $\mathrm{s}, a$ is measured in $\frac{\frac{s}{s^2}}{}$, or in IB format, $a$ is measured in $\mathrm{m} \mathrm{s}^{-2}$.

FYI

Many textbooks use $\Delta v=v_f-v_i$ for change in velocity, $v_f$ for final velocity and $v_i$ initial velocity. IB gets away from the subscripting mess by choosing $v$ for final velocity and $u$ for initial velocity.

EXAMPLE:

A driver sees his speed is $5.0 \mathrm{~m} \mathrm{~s}^{-1}$. He then simultaneously accelerates and starts a stopwatch. At the end of 10 . s he observes his speed to be $35 \mathrm{~m} \mathrm{~s}^{-1}$. What is his acceleration?

SOLUTION:

Label each number with a letter:
$v=35 \mathrm{~m} \mathrm{~s}^{-1}, u=5.0 \mathrm{~m} \mathrm{~s}^{-1}$, and $t=10$. s.
Next, choose the formula: $a=\frac{v u}{t}$.
Now substitute and calculate:

$ a=\frac{35.5}{10}=3.0 \mathrm{~m} \mathrm{~s}^{-2} .$

PRACTICE:

Why is velocity a vector?
Why is acceleration a vector?

SOLUTION:

Velocity is a displacement over time. Since displacement is a vector, so is velocity.
Acceleration is a change in velocity over time. Since velocity is a vector, so is acceleration.

Determining instantaneous and average values for velocity, speed and acceleration

∙Consider a car whose position is changing.
∙A patrol officer is checking its speed with a radar gun as shown.
∙The radar gun measures the position of the car during each successive snapshot, shown in yellow.
∙How can you tell that the car is speeding up?
∙What are you assuming about the radar gun time?

Average speed

  • We can label each position with an $x$ and the time interval between each $x$ with a $\Delta t$.
  • Then $v_A=\frac{x_2-x_1}{\Delta t}, v_B=\frac{x_3-x_2}{\Delta t}$, and finally $v_C=\frac{x_4-x_3}{\Delta t}$.
  • Focus on the interval from $x_2$ to $x_3$.
  • Note that the speed changed from $x_2$ to $x_3$, and so $v_B$ is NOT really the speed for that whole interval.
  • We say the $v_B$ is an average speed (as are $v_A$ and $v_C$ ).

Instantaneous velocity.

  •  If we increase the sample rate of the radar gun (make the \( \Delta t \) smaller) the positions will get closer together.
  • Thus the velocity calculation is more exact.
  • We call the limit as \( \Delta t \) approaches zero in the equation $v = \frac{\Delta x}{\Delta t}$ the instantaneous velocity.
  •  For this level of physics we will just be content with the average velocity. Limits are beyond the scope of this course. You can use the calculus notes to explore limits, and derivatives, if interested.
  • By the same reasoning, if \( \Delta t \) gets smaller in the acceleration equation, our acceleration calculation becomes more precise.

• We call the limit as \( \Delta t \) approaches zero of the equation

$a = \frac{\Delta v}{\Delta t}$ the instantaneous acceleration.

• For this level of physics we will be content with the average acceleration.

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