Q: Kinematics flashcards

Topic 18: Q: Kinematics

The constant-acceleration (suvat) equations

v = u + at s = ut + ½at² s = vt − ½at² s = ½(u + v)t v² = u² + 2as

Gradients and areas on displacement–time and velocity–time graphs

s–t graph: gradient = velocity. v–t graph: gradient = acceleration; area under it = displacement.

Variable acceleration: linking s, v and a with calculus

v = ds/dt, a = dv/dt = d²s/dt². s = ∫ v dt and v = ∫ a dt, each with a constant found from given conditions.

A ball is thrown upwards at 9.8 m s⁻¹ (g = 9.8). When does it reach the top, and how high does it go?

At the top v = 0: t = 9.8/9.8 = 1 s. Height = 9.8²/(2 × 9.8) = 4.9 m. It returns to its start after 2 s.

Projectile launched at speed u, angle α above the horizontal: the two components of motion

Horizontal: constant velocity u cos α, so x = (u cos α)t. Vertical: initial u sin α, acceleration −g, so y = (u sin α)t − ½gt².

A particle is projected horizontally at 12 m s⁻¹ from 4.9 m above level ground (g = 9.8). Where does it land?

Vertically: 4.9 = ½(9.8)t², so t = 1 s. Horizontally: 12 × 1 = 12 m from the point below its start.

A particle is projected from level ground with velocity (10i + 19.6j) m s⁻¹ (g = 9.8). Find the time of flight, range and greatest height.

19.6t − 4.9t² = 0 gives t = 4 s. Range = 10 × 4 = 40 m. Greatest height = 19.6²/(2 × 9.8) = 19.6 m.

Vector kinematics

r = position, v = dr/dt, a = dv/dt. With constant acceleration: v = u + at and r = r₀ + ut + ½at², worked component by component.

Sign of g in vertical motion

Take upwards as positive, so a = −9.8 m s⁻². Losing this sign is a common error.