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Solutions to problem 1 and 4 from physics 8.01 assignment 1. The first problem involves calculating the angular velocity of the sun and the time it takes for sunset based on the given angular size and distance. The second problem deals with dimensional analysis of velocities and the kinematics of motion, including the concept of average acceleration and the relationship between distance, area under the velocity curve, and minimum distance for a given maximum acceleration.
Typology: Exercises
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Figure 1: sunset
We treat the sun as an object that is rotating about the earth with uniform angular velocity, and that rises and sets vertically (this is assured by being on the equator on the equinox). We first need to know the angle sun must rotate through during the sunset (α). If we can then get the angular speed of the sun as we see it we can get the time it takes for the sunset. α is small ⇒ α tan α = DR
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D = 13. 92 × 108 m R = 1. 50 × 1011 m α = 9. 28 × 10 −^3 radians
The angular velocity follows from fact sun will rotate 2π radians during 24 hours: ωsun = (^24) ×^260 π× 60 = 7. 27 × 10 −^5 radianss The time is the angular size of the sun divided by the angular velocity, analogous to the case with distance and linear velocity. Tsunset = (^) ωsunα = 9.^28 ×^10 − 3
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a)
Figure 2: Problem 3
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b) The Person who spends less time walking and more time run- ning will win. Initially they walk along together, but Tim starts running sooner because he has only to walk 1/2 the time it will take him, not 1/2 the distance which makes Rick take longer to reach D as the graph shows (note the specific times designated on the graph).
c) Rick t 1 = D/ Vw^2 (t 1 / : the time that Rick reaches D/2)
tR − t 1 = D/ Vr^2 tR = (^2) Vw VrD Vr +Vw
d) V (^) R = (^) tDR = (^) V^2 Vr +r^ VVww
e) Tim tT 2 Vr^ +^
tT 2 Vw^ =^ D tT = (^) (Vr +DVw )/ 2
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a = − 10 m/s^2 ∆t = 2. 0 g: hitting the ground r: reflection from the ground f = 0. 71 These are the three basic kinematic equations written with arbitrary time for the initial conditions (instead of the conventional t=0). Eq. 1: x(t) = x(t 0 ) + v(t′)(t − t 0 ) + 12 a(t − t 0 )^2 Eq. 2: v(t) = v(t 0 ) + a(t − t 0 ) Eq. 3: v^2 (t) − v^2 (t 0 ) = +2a[x(t) − x(t 0 )]
We should pay attention that t 0 in these equations is the initial time, not necessarily zero. So for each region as you see in the picture there are different initial conditions for v 0 and t 0. We should first write the appropriate y(t) for each region. Setting the two y(t)’s for the two balls gives us the time of the collision time tc. F orthef irstball :
Eq. 1 ⇒ tg 1 =
2 = 1. 41 s Eq. 3 ⇒ vg 1 =
200 = 14. 1 m/s vr 1 = f vg 1 = 10m/s vg 2 = −vr 1 ∧ Eq. 2 ⇒ tg 2 − tg 1 = 2 × 1010 = 2. 0
Because the time it takes that second ball hit the ground is again 1.41 and because 2.0 +1.41 = 2.0+ 1.41! The second hits the first ball when the first ball hits the ground itself.
However, if you change ∆t say 1.8 you would get a different answer. I also write the systematic way of solving this problem:
Between the first hit and the second one ball #1 is at y 1 (t) and ball 32 is at y 2 (t) y 1 (t) = 0 + 14. 1 f (t − 1 .41) − 5(t − 1 .41)^2 y 2 (t) = 10 − 5(t − 2)^2
Collision time: tc
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y 2 (tc) = y 1 (tc)
Solving this with will give you 3.41 for ˜t as expected! (But you have to be careful that your solution is in the region of validity of both y 1 and y 2.
Figure 4: Collision
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