Question

At the end of the 15th second ,what is the distance covered by the tip of the seconds hand of the clock, who's initial position is at 0? The length of the seconds hand is 16 cm​

Answer 1

Answer:

The hand moves 360 degrees in 60 seconds. So in 15 seconds it moves 90 degrees.

The tip of the hand is moving around the circumference of the circle with radius 1cm = 2pi cm.

It does this in 60 seconds so its speed is: 2 x pi x r / 60 cm per seconds = 0.105 cm per second.

Assume the hand moves from vertical (zero seconds) to 15 seconds.

That means at the start the velocity was 0.105 cm s-1 forward. At the end the velocity is 0.105 cm s-1 down. The difference between these vectors is found by arranging them from start to finish. Here you are doing a subtraction so you add the reverse of the first....

I.e. you form a triangle of 0.105 down and add 0.105 forward to it.... Since the opposite and adjacent sides of this triangle are both 0.105 then you know the angle from the origin is 45 degrees. Isocolies right angled triangle. Since the sign on one of the sides is negative the angle is - 45 degrees.

Find the size of the hypotenuse to get the magnitude of the velocity change:

h^2 = o^2 + a^2. Since o = a then h^2 = 2 x o^2

h^2 = 2 x 0.105^2 = 0.02205. h = 0.148.

So the magnitude of the velocity change is 0.148 cm per second.

So the answer is 0.148 cm per second at -45 degrees to the horizontal + whatever angle the hand actually started at.

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The other way of thinking about the question is in terms of angular velocity.

w = 2 pi / T where T is the period or time to do 1 revolution.

i.e. w = 2 x pi / 60 = 0.104 radians per second. Since angular velocity has only two directions, clockwise and anticlockwise, then there is no change in velocity. because the magnitude is the same.