The minute hand of a clock is 6 centimeters long and the hour hand is 2 centimeters long.?

Question

At 4pm, how far is the tip of the minute hand from the tip of the hour hand?

Answer 1

use the cosine rule
let b = 6, c= 2 and A = 120° is the angle between the hands
so a = ? is the distance between the tips
then
a² = b² + c² - 2bcCosA
= 6² + 2² - 2*6*2Cos120
= 6² + 2² - 2*6*2*(-1/2)
= 52
so
a = √52
is the distance between the tips

note: Cos120 = -cos60 = -1/2
.

Answer 2

Method 1: Using vectors: First note that the minute hand rotates at a speed of 1 rotation per hour. The circumference covered by the tip of the minute hand is 2*pi*10 centimeters, so the speed of the tip is 2*pi*10 cm/hour, or 20*pi cm/hour. At the 5pm position, the minute hand is right at the top, and its instantaneous velocity is 20*pi cm/hour purely in the positive x-axis position. Its instantaneous velocity (v1, v2) is thus: Minute hand velocity at 5pm: (v1, v2) = (20*pi, 0) For the hour hand, this rotates at 1/12 rotations per hour. The circumference covered is 2*pi*5, so the speed is (2*pi*5)/12 cm/hour, or 5*pi/6 cm/hour. At the 5pm position, by geometry we see the hour hand makes a -60 degree angle with the x axis, and its instantaneous velocity (a1, a2) is split into two coordinates: Hour hand velocity at 5pm: (a1, a2) = (5*pi/6)*(-cos(30), -sin(30)) = (5*pi/6)*(-1/2, -sqrt(3)/2) %%%%In between percent signs is edited Oops, I just interchanged cos(30) and sin(30). The above vector should be flipped, which also affects the velocity difference vector below...ugh, two independent errors, although this one was just a typo... %%%% velocity difference vector is: (v1-a1, v2-a2) = (20*pi - (-1/2)*5*pi/6, 0 - -5*(pi/6)*sqrt(3)/2) = (20*pi + 5*pi/12, 5*pi*sqrt(3)/12) %%% The above vector should really be (20*pi + 5*pi*sqrt(3)/12, 5*pi/12) %%% *** Then the rate at which the distance is changing is given by the square root of the sum of squares of the components of the velocity difference vector. *** %%%%This part in between percent signs is edited: Oops, woke up this morning realizing that my sentence above "Then the rate at which the distance is changing is given by the square root..." is completely wrong. You can use the velocity difference vector, but you also have to account for the current distance between the tips! If d(t) is the distance as a function of time then: d(t) = sqrt{x(t)^2 + y(t)^2} where x(t) and y(t) are the differences between x and y coords. Then: d'(t) = (1/d(t))[x(t)x'(t) + y(t)y'(t)] So d'(0) = (1/d(0))[x(0)(20*pi + 5*pi*sqrt(3)/12) + y(0)5*pi/12] where x(0) is the current difference between the x coords, y(0) is the current distance between y coords, and d(0) = sqrt{x(0)^2 + y(0)^2}. I see now there are a lot of answers to this question, perhaps someone already corrected my mistake. Hopefully there was no mistake in method 2. Yep, I just checked and Indica noticed my mistake above. I'll give Indica a thumbs up! =) x(0) = -2.5, y(0) = 14.330127, d(0) = 14.5465645. Yep, the answer comes out to -9.898534, same as Indica's answer. Please vote Indica for "best answer." %%%% ***** Method 2: Writing the equation of the clock Let the origin be at the middle of the clock. Let's write the equations for (x(t), y(t)) for the minute hand, and (a(t), b(t)) for the hour hand, with time t=0 being 5pm. For the minute hand: x(t) = 10*cos(-2*pi*t + pi/2) (in centimeters) y(t) = 10*sin(-2*pi*t + pi/2) (in centimeters) where the distance units are in centimeters, and the time units are in hours. Note that every hour, this rotates 2*pi radians, and I have the "minus" sign in -2*pi*t because the clock rotates clockwise, rather than counter-clockwise. I add pi/2 because that is the position of the minute hand at 5pm. Then for the hour hand: a(t) = 5*cos(-2*pi*(1/12)*t - pi/3) b(t) = 5*sin(-2*pi*(1/12)*t - pi/3) where I added the (1/12) to account for the fact that the hour hand rotates 12 times slower, and the "-pi/3" is the initial point of the hour hand at 5pm. Then distance as a function of t is: d(t) = sqrt[(x(t) - a(t))^2 + (y(t) - b(t))^2] The problem then asks to find the derivative d'(t) at t=0. d'(t) = (1/2)[(x(t) - a(t))^2 + (y(t) - b(t))^2]^{-1/2} * [2(x(t)-a(t))(x'(t) - a'(t)) + 2(y(t) - b(t))(y'(t)-b'(t))] You can calculate a'(t) and b'(t) by differentiating the cos() and sin() functions. Then plug in t=0 to find d'(0). The actual plugging in looks messy so I skip that. =) Unless I have made any errors, it should give the same answer as "Method 1."

Answer 3

Angle between hour and minute hand = 120 degrees

Therefore distance between them (say x cm) is:

a^2 = b^2 + c^2 -(2bcCosA)

= 6^2 + 2^2 - (2 x 2 x 6 x Cos120)

= 40 - (-9sqrt3)

= 40 + 9rt3 cm

Answer 4

the angle in btw should be 360 x 4/12 = 120.
using Cosine rule you should be able to determine the ans

Answer 5

I agree with the above. Just apply the formulae you were given in class, and you should be ok.
.