In one line
Write the position as x(t) and y(t); differentiate each to get velocity and acceleration components, then combine them.
Explained simply
A ball thrown across a field moves forward and up/down at the same time. If you watch its shadow on the ground and its height separately, each is a simple straight-line motion. Combine them and you get the curved path.
Technically, the position vector \( \vec{r} = x\hat{i} + y\hat{j} \); velocity \( \vec{v} = \dot{x}\hat{i} + \dot{y}\hat{j} \) is always tangent to the path; acceleration \( \vec{a} = \ddot{x}\hat{i} + \ddot{y}\hat{j} \) is generally not tangent.
How it works
- Write x(t) and y(t) (given or from the problem, e.g. projectile).
- \( v_x = dx/dt \), \( v_y = dy/dt \); \( v = \sqrt{v_x^2 + v_y^2} \), direction \( \tan^{-1}(v_y/v_x) \).
- \( a_x = dv_x/dt \), \( a_y = dv_y/dt \); \( a = \sqrt{a_x^2 + a_y^2} \).
- If the path equation y = f(x) is given with \( v_x \), use \( v_y = f'(x)\,v_x \).
- Eliminate t to get the path equation if asked.
Key formula
\( v = \sqrt{\dot{x}^2 + \dot{y}^2} \), \( a = \sqrt{\ddot{x}^2 + \ddot{y}^2} \); x, y in m, t in s.
Quick example
\( x = 2t^2 \), \( y = t^3 \) (m). At t = 1 s: \( v_x = 4t = 4 \), \( v_y = 3t^2 = 3 \), so v = 5 m/s at 36.87° to x. \( a_x = 4 \), \( a_y = 6t = 6 \), so \( a = \sqrt{52} = 7.21 \) m/s\(^2\) at 56.3° to x, clearly not along v.
Common mistakes
- Assuming acceleration is along the velocity on a curved path.
- Adding magnitudes of components instead of using the square root of the sum of squares.
- Forgetting the angle when the question asks for velocity 'in magnitude and direction'.
How MU asks it
5 to 6-mark numericals: 'The position of a particle is given by \( x = \ldots, y = \ldots \); find velocity and acceleration at t = 2 s', often followed by 'find the tangential and normal components and the radius of curvature'. Present components in a small table, then magnitudes and directions.