1-In astronomy, when measuring large distances, the astronomical unit (AU), the

Question

1-In astronomy, when measuring large distances, the astronomical unit (AU), the parsec (pc), and the light- year (LY) are often used. The AU is the mean distance from the center of the Earth to the center of the Sun; 1 AU = 1.496 × 1011 m. The pc is the distance at which 1 AU subtends an angle of exactly one second of arc (see figure). 3600 seconds of arc is equal to one degree (i.e. 3600 = 1). One LY is the distance that light travels in one year. The speed of light is approximately 300,000 km/ sec.

(a) Express the pc in AU.

(b) Express the pc in LY

(c) Express the pc in m.

2-The position, x (in meters), of an object that moves along a straight line is changing with time (seconds) as follows:

x = 16 12t + 2t2
(a) Neatly draw a plot of position x(t) vs. time for 0 t 6 sec.

(b) Neatly draw a plot of the velocity v(t) vs. time for 0 t 6 sec.
(c) Neatly draw a plot of the acceleration a(t) vs. time for 0 t 6 sec.

(d) Computethevelocityoftheobjectatt=0,t=2andt=4sec. (e) Compute the object’s acceleration at t = 0, t = 2 and t = 4 sec.

(f) When is the velocity zero, and at what position does this occur? (g) Compute the average velocity between t = 1 and t = 3.
(h) Compute the average velocity between t = 0 and t = 6.

(i) Compute the average speed between t = 0 and t = 6. (j) At what time, if any, does the object reverse direction?

Suppose you throw a stone straight up with an initial speed of 15.0 m/s. This is a one-dimensional problem.

(a) If you throw a second stone straight up 1.00 s after the first, with what speed must you throw the

3-second stone if it is to hit the first at a height of 10.0 m? (There are two answers. Are both plausible?)

(b) If you throw the second stone 1.30 s after the first, with what speed must you throw the second stone

if it is to hit the first at a height of 11.0m?

Explanation / Answer

(2) The position of an object that moves along a straight line is changing with time as follows :

x = 16 12t + 2t2                                                                   { eq.1 }

differentiating an above eq & we get

v = dx / dt = - 12 + 4t                                                             { eq.2 }

Again, differentiating an eq.2 & we get

a = dv / dt = 4 m/s2

(d) To compute the velocity of an object -

At t=0, we have

v = - 12 + 4t = [(-12) + 4 (0)]

v = 0 m/s

At t=2 sec, we have

v = - 12 + 4t = [(-12) + 4 (2 s)]

v = - 4 m/s

At t=4 sec, we have

v = - 12 + 4t = [(-12) + 4 (4 s)]

v = 4 m/s

(f) When is the velocity zero and at what position does this occur?

v = dx / dt = - 12 + 4t     

(0 m/s) = - 12 + 4t

t = 3 sec

Inserting the value of 't' in eq.1, we get

x = 16 12t + 2t2          

x = [16 12 (3 s) + 2 (3 s)2] m

x = - 2 m

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