describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction

Equations of Motion

Objective

Describe and explain the motion of a particle that moves with a uniform velocity in one direction while simultaneously experiencing a uniform acceleration in a direction perpendicular to that velocity.

Key Concepts

• Uniform (constant) velocity means zero acceleration in that direction.
• Uniform (constant) acceleration means the acceleration vector does not change with time.
• When the velocity and acceleration are perpendicular, the motion can be treated as two independent one‑dimensional motions.
• This situation is the basis of projectile motion in a uniform gravitational field (ignoring air resistance).

Coordinate System

Choose orthogonal axes:

• x‑axis: direction of the uniform velocity \$u_x\$ (horizontal).
• y‑axis: direction of the uniform acceleration \$a_y\$ (vertical, usually downwards).
• Initial position: \$(x0,\;y0)\$ at \$t=0\$.

Mathematical Description

Because the motions in the \$x\$ and \$y\$ directions are independent, we can write the equations of motion for each axis separately.

Horizontal (uniform velocity):

\$x = x0 + ux t\$

Vertical (uniform acceleration):

\$y = y0 + uy t + \tfrac{1}{2} a_y t^{2}\$

Here \$uy\$ is the initial vertical component of velocity. If the particle is launched horizontally, \$uy = 0\$.

Eliminating Time

From the horizontal equation we have \$t = \dfrac{x - x0}{ux}\$. Substituting this into the vertical equation gives the trajectory equation:

\$\$

y = y0 + \frac{uy}{ux}(x - x0) + \frac{ay}{2ux^{2}}(x - x_0)^{2}

\$\$

This is the equation of a parabola, the characteristic shape of projectile motion.

Table of Symbols

Derivation of Key Results

1. Write the independent equations for each axis (as above).
2. Solve the horizontal equation for \$t\$.
3. Substitute this expression for \$t\$ into the vertical equation.
4. Collect terms to obtain the quadratic relationship between \$y\$ and \$x\$.

Example Problem

Problem: A ball is projected horizontally from the edge of a cliff 80 m high with a speed of 12 m s⁻¹. Determine (a) the time taken to reach the ground, (b) the horizontal distance travelled, and (c) the speed just before impact.

1. Given: \$ux = 12\;\text{m s}^{-1}\$, \$uy = 0\$, \$ay = g = 9.81\;\text{m s}^{-2}\$, \$y0 = 80\;\text{m}\$, \$y = 0\$ (ground level).
2. (a) Time of flight: Use the vertical motion equation with \$y = 0\$:

   \$0 = 80 + 0\cdot t - \tfrac{1}{2} g t^{2}\$

   \$\tfrac{1}{2} g t^{2} = 80\$

   \$t = \sqrt{\frac{2\times80}{g}} \approx \sqrt{\frac{160}{9.81}} \approx 4.04\;\text{s}\$

3. (b) Horizontal distance: \$x = u_x t = 12 \times 4.04 \approx 48.5\;\text{m}\$.
4. (c) Speed just before impact:

   • Vertical velocity: \$vy = uy + a_y t = 0 + 9.81 \times 4.04 \approx 39.6\;\text{m s}^{-1}\$ (downwards).
   • Horizontal velocity remains \$vx = ux = 12\;\text{m s}^{-1}\$.
   • Resultant speed: \$v = \sqrt{vx^{2} + vy^{2}} = \sqrt{12^{2} + 39.6^{2}} \approx 41.3\;\text{m s}^{-1}\$.

Common Misconceptions

• Assuming the acceleration also acts in the direction of the uniform velocity; in reality, the acceleration is perpendicular, so it does not change the speed in that direction.
• Neglecting the initial vertical component \$u_y\$ when the launch is not perfectly horizontal.
• Confusing the trajectory equation (a parabola) with a straight line; the parabola arises because the vertical displacement depends on \$t^{2}\$ while the horizontal displacement depends linearly on \$t\$.

Suggested Diagram

Summary

When a particle moves with a constant velocity in one direction and a constant acceleration in a perpendicular direction, its motion can be described by two independent one‑dimensional equations. Eliminating the time variable yields a quadratic (parabolic) relationship between the two coordinates. This framework underlies the analysis of projectile motion, a central topic in the Cambridge A‑Level Physics syllabus.