# Applied Mathematics

**Topics:**Velocity, Mathematics, Radius of curvature

**Pages:**2 (365 words)

**Published:**September 25, 2011

TOEGEPASTE WISKUNDE 1B APPLIED MATHEMATICS 1B

1. P.4.3.1 2. P.4.3.2 3. P.4.3.3 Vraag 4 Vir die massa in P.4.3.3 by t = 3sek. bereken: (a) Die hoekspoed van die massa om die oorsprong. (b) Die krommingstraal van die boom. (c) Die koordinate van die krommingsmiddelpunt. Question 4 For the mass in P.4.3.3 at t = 3sec. calculate: (a) The angular velocity of the mass around the origin. (b) The radius of curvature of the trajectory. (c) The coordinates of the centre of curvature.

(d) Die hoekspoed om die krommingspunt.

(d) The angular velocity around the centre of curvature.

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OPLOSSINGS VIR OEFENKLAS 4 SOLUTIONS TO TUTORIAL 4

P.4.3.1 vx = v0 cos θ, vy = v0 sin θ − gt x = v0 t cos θ 1 y = v0 t sin θ − gt2 2 (a) The projectile moves up when vy is positive and down when vy is negative. vy = v0 sin θ − gt = 0 when t = The projectile is at ground level when 2v0 sin θ 1 y = 0 = t v0 sin θ − gt ⇒ t = 0 or t = 2 g Thus for t ∈ 0, v0 sin θ the projectile is moving up and for t ∈ g projectile is moving down. at t = t0 = 0, vy = v0 sin θ at t = 2v0 sin θ 2v0 sin θ , vy = v0 sin θ − g g g = −v0 sin θ v0 sin θ 2v0 sin θ , g g

v0 sin θ = ttop g

the

The velocity in the y direction is of the same size as the initial velocity but in the opposite direction (down as opposed to upwards). (b) v = v0 cos θˆ + (v0 sin θ − gt) y ¯ x ˆ |¯| = v(t) = v = 2 2 v0 cos2 θ + v0 sin2 θ − 2v0 sin θgt + g 2 t2 2 v0 − 2gv0 t sin θ + g 2 t2

(c) When the projectile hits the ground, y = 0 thus t= The distance from the origin is x= 2 2v0 sin θ cos θ g

2v0 sin θ g

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The maximum displacement is found when dx dθ dx dθ = 0, = = d2 x

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