I'm having trouble with these 2 questions. I've tried them and come up with solutions but would like someone to check if I have done them properly.(adsbygoogle = window.adsbygoogle || []).push({});

A traditional watch has a second hand 1.5 cm long from the center to the tip.

a) What is the speed of the tip of the second hand?

d = 2Pir

d = 2(3.14)(1.5)

v = d/t

v = 2(3.14)(1.5)/60s

v = 0.157 cm/s

b) what is the velocity of the tip at 15s? 45s? 60?

0.157 cm/s (right)

0.157 cm/s (left)

0.157 cm/s (up)

c) What is the change in velocity between 30s and 45s?

I'm having trouble with this one. Do I use the pythagorean theorem because one is pointing to the left and one is pointing down.

0.157^2 + 0.157^2 - sq root

v = 0.222 cm/s (sw)

A student tosses their pen vertically upward with an intial velocity of 3.8 m/s.

a) Maximum height

Vf^2 = Vi^2 + 2ad

0 = 3.8^2 + 2(-9.8)d

d = 0.736735 m

b) How much time will pass before the pen returns to his hand if it is at the same level that he released the pen at?

d = v2t - 1/2at^2

0= (3.8)(t) - 1/2(9.8)t^2

t = 0.775510 s

c) if the student's hand is 1.5 m above the ground and he misses the pen, with what speed will the pen hit the ground.

0.736735 + 1.5 = 2.236735 m

Vf^2 = Vi^2 + 2ad

Vf^2 = (0) + 2(9.8)(2.236735)

vf = 6.62117 m/s (down)

Can someone plz check these over. I would really appreciate it. Thank you.

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# Acceleration due to gravity/velocity questions (I've done the work)

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