Week 2 Homework Solutions PDF

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1. A gazelle is running in a straight line along the x-axis. The graph in the figure shows this animal’s velocity as a function of time. During the first 12.0 s, find

(a) the total distance moved Here, we need to recognize that the distance travelled is the area under a velocity curve. We have a trapezoid between 0s and 10s, whose area works out to be 80m, and a triangle from 10s-12s whose area happens to be 12m. The total area under our curve is thus 92m, and this is the distance the gazelle travels. (b) the displacement of the gazelle. Since the gazelle never ran backwards (never had a negative velocity), the displacement is the same as the distance travelled. So the answer is the same as for part (a): 92m (c) Sketch an ax -t graph showing this gazelles acceleration as a function of time for the first 12.0 s. The acceleration can be determined from the slope of the v-t graph: For the first 10 s we have: acceleration = rise/run = (6 m/s) / (10 s) = +0.6 m/s2 For the final 2 s we have: acceleration = rise/run = (-12 m/s) / (2 s) = -6 m/s2

ax [m/s2]

2 t

-6

Problem 2.15 Position of turtle as a function of time: 𝑥(𝑡) = 50.0𝑐𝑚 + (2.00𝑐𝑚/𝑠)𝑡 − (0.0625𝑐𝑚/𝑠2 )𝑡 2 a) There are two approaches to finding the turtle’s initial velocity, initial position, and acceleration.

Approach 1: We have the expression for the position of the turtle: 𝑥(𝑡) = 50.0𝑐𝑚 + (2.00𝑐𝑚/𝑠)𝑡 − (0.0625𝑐𝑚/𝑠2 )𝑡 2

To find the velocity of the turtle, we take the derivative of position with respect to time: 𝑣(𝑡) =

𝑑𝑥 𝑑𝑡

= (2.00𝑐𝑚/𝑠) − (0.0625𝑐𝑚/𝑠2 )(2𝑡)

To find the acceleration, we take the derivative of velocity with respect to time: 𝑎(𝑡) =

𝑑𝑣 = −(0.0625𝑐𝑚/𝑠 2 )(2) 𝑑𝑡

To find the initial values of these quantities, we set t = 0

Initial Position: Initial Velocity:

𝑥(𝑡 = 0) = 50.0𝑐𝑚

𝑣(𝑡 = 0) = 2.00𝑐𝑚/𝑠

Notice that the expression for a(t) has no time dependence, I.e. the turtle’s acceleration is constant: 𝑎(𝑡) = −0.125𝑐𝑚/𝑠 2

Approach 2: Notice that the equation for x(t) has the same form as equation 2.12 in the text: 𝑥(𝑡) = 𝑥𝑖

+

𝑣𝑖 𝑡

+

1 2 𝑎 𝑡 2

𝑥(𝑡) = 50.0𝑐𝑚 + (2.00𝑐𝑚/𝑠)𝑡 − (0.0625𝑐𝑚/𝑠2 )𝑡 2 This allows us to identify the terms: 𝑥𝑖 = 50.0𝑐𝑚 𝑣𝑖 = 2.00𝑐𝑚/𝑠 1

2

𝑐𝑚

⟹ 𝑎 = −0.125𝑐𝑚/𝑠 2

𝑎 = −0.0625 𝑠2

b) Given the position as a function of time we can find the velocity by differentiating: 1 𝑥(𝑡) = 𝑥𝑖 + 𝑣𝑖 𝑡 + 𝑎𝑡 2 2 𝑣(𝑡) =

𝑑𝑥 = 𝑣𝑖 + 𝑎𝑡 𝑑𝑡

Note that this is equation 2.8 from the text.

We want the time at which v = 0, so set v = 0 and solve for t: 0 = 𝑣𝑖 + 𝑎𝑡 ⇒𝑡=−

𝑣𝑖 𝑎

Substituting the numerical values, this gives: 𝑡=

2.00𝑐𝑚/𝑠

0.125𝑐𝑚/𝑠 2

= 16𝑠

c) We want the time at which x = xi , so set x = xi in our position equation and solve for t: 1 𝑥𝑖 = 𝑥𝑖 + 𝑣𝑖 𝑡 + 𝑎𝑡 2 2 ⇒

1 𝑣𝑖 𝑡 = − 𝑎𝑡 2 2

There are two solutions: t=0

𝑡=−

2𝑣𝑖 𝑎

We take the second solution, since the question asks for the time at which the turtle returns to its initial position. Substituting numerical values we have: 2.00𝑐𝑚 2( 𝑠 ) 𝑡=− = 32𝑠 0.125𝑐𝑚 (− ) 2 𝑠 d) We want to find the times at which the turtle is 10 cm from its starting point, i.e. when the displacement of the turtle Δ𝑥 = ±10𝑐𝑚 The displacement is given by: Δ𝑥 = 𝑥𝑓 − 𝑥𝑖 = 𝑣𝑖 𝑡 +

1 2 𝑎𝑡 2

This is a quadratic equation: 1 2 𝑎𝑡 + 𝑣𝑖 𝑡 − Δ𝑥 = 0 2

⇒

1 −𝑣𝑖 ± √𝑣𝑖2 − 4 ( 𝑎) (−∆𝑥) 2 𝑡= 1 2 ( 2 𝑎)

⇒

𝑡=

−𝑣𝑖 ± √𝑣𝑖2 + 2𝑎∆𝑥 𝑎

This is going to generate four solutions! We have the ± in the quadratic formula, and then both positive and negative values for Δ𝑥 = ±10𝑐𝑚 Considering first the positive displacement: Δ𝑥 = +10𝑐𝑚 ⇒

𝑡=

−2.00𝑐𝑚/𝑠 ± √(2.00𝑐𝑚/𝑠 )2 + 2(−0.125𝑐𝑚/𝑠2 )(+10𝑐𝑚) (−0.125𝑐𝑚/𝑠2 )

t = 6.2 s

or

t = 25.8 s

then the negative displacement: Δ𝑥 = −10𝑐𝑚 ⇒

𝑡=

−2.00𝑐𝑚/𝑠 ± √(2.00𝑐𝑚/𝑠 )2 + 2(−0.125𝑐𝑚/𝑠2 )(−10𝑐𝑚) (−0.125𝑐𝑚/𝑠2 )

t = -4.4 s

or

t = 36.4 s

Probably we should discard the negative solution. The equations of kinematics do not know or care about the direction of time. As far as they are oncerned, the turtle has been crawling that straight line since the universe began, and will continue to do so till it ends. We should probably assume that the turtle began its straight line journey at t = 0, so any times before that are not valid.

To determine the velocity at these times we use: 𝑣(𝑡) = 𝑣𝑖 + 𝑎𝑡

NB. We could just insert the times calculated above into the equation, but that’d be shoddy! It could lead to rounding errors, will only give answers specific to this particular turtle, and completely fails to illuminate the relationships between the kinematic quantities. Instead, we will insert the algebraic solution for the time.

⇒

𝑣(𝑡) = 𝑣𝑖 + 𝑎

⇒

NB. This is just a version of:

−𝑣𝑖 ± √𝑣𝑖2 + 2𝑎∆𝑥 𝑎

𝑣(𝑡) = ±√𝑣𝑖2 + 2𝑎∆𝑥

𝑣𝑓2 − 𝑣𝑖2 = 2𝑎∆𝑥

Considering first the positive displacement: Δ𝑥 = −10𝑐𝑚 𝑣(𝑡) = ±√(2.00𝑐𝑚/𝑠 )2 + 2(−0.125𝑐𝑚/𝑠2 )(+10𝑐𝑚)

v = +1.22 m/s

or

v = -1.22 m/s

then the negative displacement: Δ𝑥 = −10𝑐𝑚 𝑣(𝑡) = ±√(2.00𝑐𝑚/𝑠 )2 + 2(−0.125𝑐𝑚/𝑠2 )(−10𝑐𝑚)

v = +2.55 m/s

or

v = -2.55 m/s

NB. We will ignore the solution corresponding the negative time.

e) The position of the turtle is quadratic in time: 1 𝑥𝑖 = 𝑥𝑖 + 𝑣𝑖 𝑡 + 𝑎𝑡 2 2

𝑥(𝑡) = 50.0𝑐𝑚 + (2.00𝑐𝑚/𝑠)𝑡 − (0.0625𝑐𝑚/𝑠2 )𝑡 2 Initially the turtle moves further from the origin, but then changes direction and moves in the negative x direction.

The velocity of the turtle is linear in time: 𝑣(𝑡) = 𝑣𝑖 + 𝑎𝑡

𝑣(𝑡) = (2.00𝑐𝑚/𝑠) − (0.125𝑐𝑚/𝑠2 )𝑡

The acceleration of the turtle is constant in time: 𝑎(𝑡) = −0.125𝑐𝑚/𝑠 2

3. A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a height H, directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. We should write expressions for the position of both balls in terms of time, set them equal, and find the time that satisfies the equation. I.e. Find ∆𝑡 when y1f = y2f 1

∆𝑦 = 𝑣𝑖 ∆𝑡 + 2 𝑎(∆𝑡)2 1

𝑦𝑓 = 𝑦𝑖 + 𝑣𝑖 ∆𝑡 + 2 𝑎(∆𝑡)2 1

Ball 1: 𝑦1𝑓 = 0 + 𝑣0 ∆𝑡 + 2 (−𝑔)(∆𝑡)2

since ball is initially at the ground (yi = 0) and is thrown (vi=v0)

1

Ball 2: 𝑦2𝑓 = 𝐻 + (0)∆𝑡 + 2 (−𝑔)(∆𝑡)2 since ball is initially at height H (yi = H) and is dropped (vi=0) The balls collide when 𝑦1𝑓 = 𝑦2𝑓 1

1

𝑣0 ∆𝑡 + 2 (−𝑔)(∆𝑡)2 = 𝐻 − 2 𝑔(∆𝑡)2 (b) Find the value of H in terms of v0 and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

4. At ti = 0 a ball is thrown straight upward from yi = 0 at an initial speed vi . While in the air it is in free fall.

(a) Sketch ay -t, vy -t and y-t graphs for the motion. For the plots below, I chose vi = 5 m/s. The exact details don’t matter; the qualitative takeaways are that: • The acceleration graph is constant. • The velocity graph decreases linearly, and it corresponds to the slope of the position graph. As a result, the zero point of the velocity graph corresponds to the maximum of the position graph. • The position graph is parabolic.

(b) What are the magnitude and direction of the acceleration while the boulder is • Moving upward? • Moving downward? • At the highest point? The acceleration is down, with magnitude g = 9.8m/s2 , for all of these situations.

(c) When is the velocity of the ball zero? The velocity of the ball is zero at the top of its arc, or when it has lost all of its initial vertical velocity to gravity: v =vi − gt 0 =vi − gt t =vi /g

(d) Find the maximum height ymax reached by the ball. The ball is at the top of its arc when its velocity is zero, aka at the time we found in (c). So its position is 1 2 gt + vi t 2    2 vi 1 vi 1 vi2 + vi =− g = g 2 g 2 g

y =− ymax

(e) Find the time at which the ball reaches this maximum height. We should recognize that this is the same question as (c), and copy our answer t = vi /g

(f) At what time is the ball moving at vi /2 upward? Following (c): v =vi − gt vi /2 =vi − gt vi 0 = − gt 2 vi t= 2g

(g) At what time is the ball moving at vi /2 downward? More of the same: v =vi − gt −vi /2 =vi − gt 3vi − gt 0= 2 3vi t= 2g

(h) At what time does the ball return to its initial position? We can solve for t explicitly: 1 2 gt + vi t 2 1 2 0 = − gt + vi t 2 p vi ∓ vi vi −vi ± v2i = t= = 0, 2 g g 2(− 12 g) y =−

Obviously the t = 0 solution corresponds to our start time, so the answer is the nonzero solution t =

2vi . g

Alternatively, we can recognize that the ball spends half its time rising and half its time falling, so our answer twice of either (c) or (e), which is indeed the case. Yet another way to do this problem is to recognize that the ball will hit the ground with velocity −vi . Knowing this, we just need to compute the time required to effect this change in velocity: ∆v = vf − vi = −vi − vi = −2vi 2vi −2vi = t= −g g

(i) How long is the ball in the air above some arbitrary height h? In other words, when is the ball at height h, and what is the time between those points? 1 2 gt + vi t 2 1 0 = − gt2 + vi t − h 2 q   −vi ± vi2 − 4 − 21 g (−h)   t= 2 − 21 g p vi ∓ vi2 − 2gh t= g

h=−

Cleaning this up, we denote t+ as the larger solution and t− as the smaller: p vi + vi2 − 2gh t+ = g p vi − vi2 − 2gh t− = g The time between them is the time the ball spends above h: p 2 vi2 − 2gh t+ − t− = g

(j) Check that your answer for (i) makes sense. If you set the height to h = 0, you should get the same answer as for part (h). If you set the height to h = ymax you should get ∆t = 0. Setting h = 0: t+ − t− = That’s correct. Setting h = ymax =

2vi g

1 v2i : 2g

t+ − t− =

2

p

v2i − 2gh 2 = g

p

vi2 − 2g(vi2 /2g) =0 g

So we’re done.

5. An electric fan is turned off, and its angular velocity decreases uniformly from ωi to ωf in the time interval ∆t.

(a) Find the angular acceleration The angular acceleration is exactly analogous to linear acceleration, so ∆ω ωf − ωi = ∆t ∆t

α=

(b) Find the number of revolutions made by the motor in the time interval ∆t The angular distance is analogous to linear distance: 1 θ(∆t) = α(∆t)2 + ωi ∆t + θ0 2 1 ωf − ωi (∆t)2 + ωi ∆t + θ0 = 2 ∆t 1 = (ωf + ωi )∆t + θ0 2 The distance the motor travels is θ(∆t) − θ (0): 1 ∆θ = (ωf + ωi )∆t 2 The number of revolutions is this angular distance divided by the “distance” of one revolution, 2π : Nrev =

(ωf + ωi )∆t ∆θ = 2π 4π...