PHYS2250 Chap.2 Kinematics and Projectile Motion

Kinematics and Projectile Motion

Translate among equations, graphs and trajectories

Kinematics connects position, velocity and acceleration through derivatives and integrals. Motion graphs reveal slopes and areas, while projectile motion separates horizontal and vertical components linked by one time variable.

The chapter distinguishes instantaneous from average quantities, turning points from equilibrium, constant from variable acceleration, and geometric landing constraints from memorised range formulae.

Reasoning route and exam method

Kinematics describes motion before asking what caused it.

Position is the primitive description, velocity is its time derivative, and acceleration is the derivative of velocity; this relationship belongs in the diagnostic checklist for the chapter. Those statements are vector statements. In one dimension a sign change can be read from a line; in two dimensions it must be tracked independently in each component; this is a useful checkpoint before any numerical substitution.

Constant-acceleration formulae are useful only after confirming that acceleration really is constant over the interval; the same result should survive the relevant limiting-case test.

When it is not, integration replaces memorised formulae and the initial conditions determine the constants; that distinction determines which quantities may be combined.

With uniform gravity and negligible resistance, horizontal acceleration is zero and vertical acceleration is −g; this observation fixes the physical interpretation of the algebra. Both components share the same time.

That shared clock is the link between them: solve the vertical equation for a physically admissible time, then use it in the horizontal equation; the statement supplies a direct check on the final result. At maximum height the vertical velocity is zero, but the acceleration remains downward and the horizontal velocity remains unchanged; this condition should remain visible beside the governing equation.

On an inclined landing surface the final constraint is geometric, so substitute the trajectory into the line equation rather than searching for a special projectile formula.

A ball leaves ground level at 20 m s⁻¹ and 30° above horizontal. Using g = 9.8 m s⁻², the initial components are 17.32 m s⁻¹ horizontally and 10.0 m s⁻¹ vertically. At a wall 15 m away, time is t = 15/17.32 = 0.866 s.

The height is y = 10.0(0.866) − 4.9(0.866)² = 4.98 m. Therefore a 4.5 m wall is cleared by 0.48 m. The downward term must be subtracted; using speed in place of the vertical component is the typical losing step.

Prompt. A position-time curve is horizontal at one instant and concave upward. What are the signs of velocity and acceleration?

Answer.

Velocity is zero at that instant because the tangent is horizontal; acceleration is positive because the slope is increasing.

Prompt. A formula with linear drag should recover ordinary projectile motion as the drag constant tends to zero. What should you inspect?

Answer. Expand the exponential to first nonzero order before taking the limit.

Direct substitution often creates 0/0 and hides the cancellation.

TWO EQUATIONS Component route. x = uₓt + ½aₓt² and y = uᵧt + ½aᵧt². Use one time variable and impose the physical boundary only after both component equations are written.

On a position-time graph the instantaneous velocity is the tangent slope, while a changing slope signals nonzero acceleration.

Reading the curve locally prevents average velocity from being mistaken for the instantaneous value.

A turning point has zero velocity in one component, not zero acceleration and not necessarily zero total speed.

What this chapter covers

• 01

  position velocity acceleration

• 02

  motion graph slopes

• 03

  constant acceleration

• 04

  projectile components

• 05

  inclined landing constraints

Worked example · free

Clear a wall with a projectile

Key terms

Instantaneous Velocity

The time derivative of position at one instant.

Projectile Motion

Two-dimensional motion under gravity with horizontal and vertical components sharing time.

Turning Point

An instant when one velocity component vanishes before reversing while acceleration can remain nonzero.

Kinematics and Projectile Motion FAQ

Exam move

Sketch position, velocity and acceleration graphs for the same motion. Then solve projectiles by components and verify the time satisfies both the trajectory and landing constraint. Use three contrasting cases: constant positive acceleration, a turning point under negative acceleration and a piecewise motion with a rest interval. Label slopes and signed areas rather than merely drawing shapes.

For projectiles, change the launch angle while keeping speed fixed and predict which component grows before calculating. Add one inclined landing line so that the terminal condition comes from geometry instead of ground level. After solving, substitute the computed time into both component equations and check whether the resulting position really lies on the stated boundary.

Review any discrepancy at the first mismatched equation, because a correct-looking range built from an inconsistent time is not a valid trajectory. Finish by narrating the signs of velocity and acceleration at launch, apex and landing, because that verbal pass catches component swaps that a numerical range can conceal.

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