Interactive Lecture
Motion in One Dimension
by Dr Arnaldo Delli Carri
Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion. It focuses purely on the geometry of motion - how things move, not why they move.
In kinematics, we study the relationships between:
Linear kinematics focuses on motion along a straight line in one dimension (1D).
To measure position, a reference frame is needed.
Displacement (Δs) is the change in position of an object:
\(\Delta s = s_f - s_i\)
Where:
as measured within the chosen reference frame.
SI Unit: metres (m)
Expressing the displacement as a function of time provides the function \(s(t)\), the equation of motion (also called the position-time relation).
Displacement is a vector quantity: it has both magnitude and direction. In 1D motion, direction is indicated by the sign (+ or -).
Displacement differs from distance: distance is always positive (scalar), whilst displacement can be negative.
Velocity (v) is the rate of change of position with respect to time:
Average velocity: \(v_{\text{avg}} = \displaystyle \frac{\Delta s}{\Delta t} = \frac{s_f - s_i}{t_f - t_i}\)
Instantaneous velocity: \(v(t) = \displaystyle \frac{ds(t)}{dt}\)
SI Unit: metres per second (m/s)
Velocity tells us two things:
Acceleration (a) is the rate of change of velocity with respect to time:
Average acceleration: \(a_{\text{avg}} = \displaystyle \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}\)
Instantaneous acceleration: \(a = \displaystyle \frac{dv(t)}{dt} = \frac{d^2s(t)}{dt^2}\)
SI Unit: metres per second squared (m/s²)
Depending on the sign of velocity, acceleration can have:
Observe how acceleration affects an object's motion. Click to apply acceleration!
Click and hold on the track to apply acceleration. Release to coast at constant velocity.
For constant acceleration, we have five key kinematic equations:
Where:
These equations only apply when acceleration is constant.
For varying acceleration, we need calculus methods (integration).
Set the initial velocities and accelerations for two cars and watch them race to the finish line!
| Concept | Equation | Description |
|---|---|---|
| Velocity | \(v = u + at\) | Final velocity from initial velocity and acceleration |
| Displacement (time) | \(s = ut + \frac{1}{2}at^2\) | Displacement as function of time |
| Displacement (velocity) | \(v^2 = u^2 + 2as\) | Velocity-displacement relationship |
| Average Velocity | \(v_{avg} = \frac{u + v}{2}\) | Average velocity under constant acceleration |
| Displacement (average) | \(s = v_{avg} \cdot t\) | Displacement from average velocity |
A car travelling at 30 m/s (108 km/h) needs to make an emergency stop. If the maximum deceleration is 8 m/s², calculate:
Show Solution
Given: \(u = 30\) m/s, \(v = 0\) m/s (stops), \(a = -8\) m/s²
a) Stopping distance:
Using: \(v^2 = u^2 + 2as\)
\(0 = 30^2 + 2(-8)s\)
\(0 = 900 - 16s\)
\(s = \frac{900}{16} = 56.25\) m
b) Time to stop:
Using: \(v = u + at\)
\(0 = 30 + (-8)t\)
\(t = \frac{30}{8} = 3.75\) s
Answer: stopping distance \(s = 56.25\) m; time to stop \(t = 3.75\) s.
A car travelling at 20 m/s begins to overtake a lorry. It accelerates at 2 m/s² for 5 seconds. Calculate:
Show Solution
Given: \(u = 20\) m/s, \(a = 2\) m/s², \(t = 5\) s
a) Final velocity:
Using: \(v = u + at\)
\(v = 20 + 2(5) = 30\) m/s
b) Distance covered:
Using: \(s = ut + \frac{1}{2}at^2\)
\(s = 20(5) + \frac{1}{2}(2)(5^2)\)
\(s = 100 + 25 = 125\) m
Answer: final velocity \(v = 30\) m/s; distance covered \(s = 125\) m.
A train accelerates from rest at 1.5 m/s² for 20 seconds, then travels at constant velocity for 60 seconds. Find the total distance travelled.
Show Solution
Stage 1: Acceleration
Given: \(u = 0\) m/s, \(a = 1.5\) m/s², \(t = 20\) s
Distance: \(s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(1.5)(20^2) = 300\) m
Final velocity: \(v = u + at = 0 + 1.5(20) = 30\) m/s
Stage 2: Constant velocity
Given: \(v = 30\) m/s, \(t = 60\) s
Distance: \(s_2 = vt = 30(60) = 1800\) m
Total distance: \(s_{total} = s_1 + s_2 = 300 + 1800 = 2100\) m
Answer: total distance travelled \(= 2100\) m.