# Help with trigonometry question

• Nov 28th 2010, 05:47 AM
lauangela
Help with trigonometry question
Hi, i need help in this trignometry question and i don't quite understand how to solve it this question is dealing with the equation:

y-k sin or cos (x-h)
---- = -----------------
b a

just in case your not sure

k = vertical shift
b = amplitude
h = horizontal shift
a = period

here is the question:

a tire with a diameter of 63 cm coasts at 1.6 m/s over a nail at exactly 12:00 noon. the nail is stuck in the tire but does not deflate

a. how high above the ground will the nail be at:
1. 12:00:08
2. 12:00:15
3: 12:01:00
4: 12:15:00

b. Find the first four times the nail at 10 cm above the ground

thanks! i really need to understand this question
• Nov 28th 2010, 06:18 AM
skeeter
Quote:

Originally Posted by lauangela
Hi, i need help in this trignometry question and i don't quite understand how to solve it this question is dealing with the equation:

y-k sin or cos (x-h)
---- = -----------------
b a

just in case your not sure

k = vertical shift
b = amplitude
h = horizontal shift
a = period

here is the question:

a tire with a diameter of 63 cm coasts at 1.6 m/s over a nail at exactly 12:00 noon. the nail is stuck in the tire but does not deflate

a. how high above the ground will the nail be at:
1. 12:00:08
2. 12:00:15
3: 12:01:00
4: 12:15:00

b. Find the first four times the nail at 10 cm above the ground

thanks! i really need to understand this question

the first thing you need to determine is the period of up and down motion for the nail.

$\omega = \frac{v}{r} = \frac{160 \, cm/s}{31.5 \, cm} = \frac{320}{63} \, rad/s$

period, $T = \frac{2\pi}{\omega} = \frac{63\pi}{160} \approx 1.24 \, s$

at $t = 0$, the nail is on the ground and rises to a height of 63 cm at $t = \frac{63\pi}{320} \approx 0.62 \, s$

since the cycle starts at a low point (h = 0) , goes up to a height of h = 63 cm and back down, a negative cosine curve is probably the best model to simulate the motion of the nail up and down over time.

$h = -31.5 \cos\left(\frac{320}{63} \cdot t\right) + 31.5$

where h is height above the ground in cm and t is time in seconds.
• Nov 28th 2010, 06:38 AM
lauangela
thanks for the reply but i am still very confuse about the whole thing

1. 12:00:08
2. 12:00:15
3: 12:01:00
4: 12:15:00

how high?

im sorry but i'm just confuse by the whole work

i need to identify the periods, amplitude, horizontal and vertical shift but then i can't see it and overall confuse
• Nov 28th 2010, 06:44 AM
skeeter
Quote:

Originally Posted by lauangela
thanks for the reply but i am still very confuse about the whole thing