Lesson 6 of 10 · 7 min
Velocity and acceleration in a plane
NCERT §3.7
A camera drone films the match from above. From the moment it starts, its controller logs its position as r = 4.0t î + 1.5t² ĵ metres, with t in seconds. Where is it heading at t = 1 s?
The lesson in notes
In short
The position vector is r = x î + y ĵ; as the object moves from r to r′ in Δt, the displacement is Δr = r′ − r.
Average velocity is Δr/Δt, a vector along Δr. Instantaneous velocity is the limit v = dr/dt, with components vx = dx/dt and vy = dy/dt.
The instantaneous velocity is always along the tangent to the path, in the direction of motion. Its magnitude is v = √(vx² + vy²) and its angle to x is given by tan θ = vy/vx.
Average acceleration is Δv/Δt; instantaneous acceleration is a = dv/dt, with ax = dvx/dt = d²x/dt² and ay = dvy/dt.
In one dimension velocity and acceleration lie on the same line; in a plane the angle between v and a can be anything from 0° to 180°.
Example: r = 4.0t î + 1.5t² ĵ (metres, t in s) gives v = 4.0 î + 3.0t ĵ and a = 3.0 ĵ m/s²; at t = 1 s the speed is 5.0 m/s at about 37° to the x-axis.