Given:
r = 0.02 m r=0.02\:\rm m r = 0.02 m
T = 60 s T=60\:\rm s T = 60 s
(i) the speed of the tip of the second hand
v = 2 π r T = 2 π ∗ 0.02 60 = 0.0021 m / s = 2.1 m m / s v=\frac{2\pi r}{T}=\frac{2\pi*0.02}{60}=0.0021\:\rm m/s=2.1\: mm/s v = T 2 π r = 60 2 π ∗ 0.02 = 0.0021 m/s = 2.1 mm/s
ii) the velocity of the tip at t = 0 t=0 t = 0
v 1 = 2.1 m m / s t o t h e r i g h t v_1=2.1\:\rm mm/s\quad to\: the\;right v 1 = 2.1 mm/s to the right and t = 15 t=15 t = 15 seconds
v 2 = 2.1 m m / s d o w n w a r d v_2=2.1\:\rm mm/s\quad downward v 2 = 2.1 mm/s downward iii) change in velocity from t = 0 t=0 t = 0 to t = 15 t=15 t = 15 seconds
Δ v = 2 v = 2 ∗ 2.1 m m / s = 3.0 m m / s \Delta v=\sqrt{2}v=\sqrt{2}*2.1\:\rm mm/s=3.0\:\rm mm/s Δ v = 2 v = 2 ∗ 2.1 mm/s = 3.0 mm/s iv) the average acceleration
a = Δ v Δ t = 3.0 m m / s 15 s = 0.2 m m / s 2 a=\frac{\Delta v}{\Delta t}=\frac{3.0\:\rm mm/s}{15\:\rm s}=0.2\:\rm mm/s^2 a = Δ t Δ v = 15 s 3.0 mm/s = 0.2 mm/ s 2
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