Answers/Solutions: Giancoli Chapter 3

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Q7: The maximum value of V1 + V2 occurs when the two vectors are parallel: in this case the value of the sum is 7.5 km. The minimum value occurs when they are anti-parallel (i.e., point in opposite directions). In this case the sum would be 0.5 km in the direction of V2.

Q9(a): The magnitude of a vector will equal its only non-zero component if the vector lies along one of the coordinate axes. For example, if A = 4.0i, then the magnitude A = 4.0 km, which is the same as Ax.

(b) The magnitude A is equal to the square-root of the sum of squares of the components. Thus A can never be less than any one of its components!

Q10: A particle with constant speed can nevertheless be accelerating if its direction is changing.

A particle with constant velocity cannot be accelerating, by definition: acceleration is DEFINED as the rate of change of velocity! For constant velocity, both the speed and the direction must be constant.

Q15: A projectile's speed is the magnitude of its velocity vector:

s = sqrt(vx2 + vy2)

Now the horizontal component vx does not change during the fall, while the vertical component vy does: it is positive during the projectile's rise, zero at the peak, then negative during the descent. Clearly s is a minimum when vy is zero; i.e., at the top of its trajectory.

P3: Magnitude is V = sqrt[(7.802 + (-6.40)2] = 10.1

Angle CCW from +x axis is \theta \ where tan(\theta \ ) = (-6.40/7.80). Thus tan(\theta \ ) = -0.821. Using the inverse tangent (i.e. Arctan) button on my calculator gives -39.4 degrees. But I need to check this, because the angle 140.6 degrees (i.e., -39.4 + 180) also has a tangent of -0.821 (check this on your calculator).

"Checking" means looking at the two components and seeing in which quadrant the vector lies. Since the x-component is positive and the y-component is negative, the angle must be in the fourth quadrant, and so -39.4 degrees is indeed the answer. IF the angle were 140.6 degrees, which is in the second quadrant, the x-component would be negative and the y-component positive: (-7.80, +6.40). The tangent of this angle is the same as our answer, but the two angles differ by 180 degrees. You should always check the components when using inverse tangent to find an angle since there will always be two vectors (opposite in direction, differing in angle by 180 degrees) which have the same tangent. The calculator will (usually) give you an angle between +90 and -90 degrees - that is, in the first or fourth quadrant. The answer will always be either the angle given by the calculator OR this angle plus (or minus!) 180 degrees.

P12: We are given that A has magnitude (i.e., length) 44.0 and orientation +28 degrees (i.e., 28 degrees CCW from +x axis). The vector C has magnitude 30.0 and is oriented along the -y axis (i.e., 270 degrees CCW from the +x axis). We want the vector R = A - C.

We immediately compute the x and y components of A and C. The results are

Ax = A cos(28) = (44.0)(0.883) = 38.8

Ay = A sin(28) = (44.0)(0.469) = 20.7

Bx = B cos(270) = (30.0)(0) = 0

By = B sin(270) = (30.0)(-1) = -30.0

(you can, of course, get the components of B immediately from the Figure)

Then we obtain the results for R = A - C :

Rx = Ax - Cx = 38.8 - 0 = 38.8

Ry = Ay - Cy = 20.7 - (-30) = -50.7

Finally, R = sqrt[Rx2 + Ry2] = sqrt[(38.8)2 + (-50.7)2)] = 63.8

and R makes an angle \theta \ (measured CCW from the +x axis) whose tangent is given by

tan(\theta \ ) = Ry\Rx = -50.7/38.8 = -1.307

My calculator gives \theta \ = Arctan(-1.307) = -52.6 degrees. Checking the components, we see that Rx is positive and Ry is negative, therefore the angle is in the fourth quadrant and is indeed given by -52.6 degrees.

So, R is a vector 63.8 units long pointing 52.6 degrees CW from the +x axis.

P28: The governing equation is 3-13. In component form:

      • x(t) = x0 + (vo)xt + (1/2)axt2
      • y(t) = y0 + (vo)yt + (1/2)ayt2

and, differentiating these with respect to time, we get (although we don't use them in this problem):

      • vx(t) = (vo)x + axt
      • vy(t) = (vo)y + ayt

We need to "tailor" these equations (well, just the first two in this case) to this particular problem. Take +y to be up and set +x in the direction of the leap. Then in this problem, (vo)x = +3.2 m/s, (vo)y = 0. If we set the (x,y) origin on the ground directly under her leap, then xo = 0 and yo = +7.5 m. Finally, ay = -g = -9.8 m/s2 and ax = 0.

So, plugging in these values for xo, yo, (vo)x, (vo)y, ax, and ay, we have our mathematical model to describe the motion in this particular case:

x(t) = 3.2t

y(t) = 7.5 - 4.9t2

We want x at the instant when y = 0. Call this time t1. At this instant, the equations become:

x(t1) = 3.2t1 = R

y(t1) = 7.5 - 4.9(t1)2 = 0

From the second equation, t1 = (7.5/4.9)1/2 = 1.24 sec. So:

R = (3.2)(1.24) = 3.96 m

P38: The governing equations are the four *** equations in the previous problem (again, we only need the first two). Let's take upward as the +y direction and set our origin at ground level directly underneath where the ball is hit: then xo = 0 and yo = +1.0 m. We also have (vo)x = vocos(\theta \ o) and (vo)y = vosin(\theta \ o). Finally, we know that ax = 0 and ay = -g = -9.8 m/s2.

Plugging these values into the governing equations yields:

x(t) = (27.0)cos(45)t = (19.1)t

y(t) = (27.0)sin(45)t - (4.9)t2 = (19.1)t - (4.9)t2

The question (translated into mathspeak) is: what is the value of x when y = 13.0 m? Call the time at this point t1. So at time t1, we have

x(t1) = (19.1)t1 = R

y(t1) = (19.1)t1 - (4.9)t12 = 13.0

Two equations, two unknowns (R and t1). We solve the second equation for t1. The quadratic formula gives (as usual) two solutions: t1 = 0.88 sec and t1 = 3.02 sec. Clearly the shorter time corresponds to when the ball reaches roof level (13 m) on its way up, but we want the time it hits the roof on a downward path, so we use t1 = 3.02 seconds. Using this value in the equation for x gives x(3.02) = 57.7 m.

P45: NOW we will need all four of the set of *** equations in P28 above.

Make a coordinate system: let (0,0) be the point of take-off and upward by the +y direction. Call the instant of splashdown t1. Then xo = yo = 0. Also ax = 0 and ay = -g = -9.8 m/s2. Finally x(t1) = +3 and y(t1) = -5.

(a) In the first two equations, then, the unknowns are the x and y components of vo. Plugging in the data at the instant t1 = 1.3 seconds:

+3 = (vo)x(1.3)

-5 = (vo)y(1.3) - (4.9)(1.3)2.

Solutions are

(vo)x = 2.31 m/s

(vo)y = 2.52 m/s

So vo = sqrt[(2.312 + (2.52)2] = 3.42 m/s at a angle of \theta \ = Arctan (2.52/2.31) = 47.5 degrees (in the first quadrant since both components are positive).

(b) We are seeking the value of y when vy = 0. Call this time t2. We will solve the fourth *** equation in P28 for t2, then use that value of time to find y(t2) in the second *** equation. So:

0 = 2.52 - (9.8)t2

which yields t2 = 0.257 seconds. At that instant

y(0.257) = (2.52)(0.257) + (0.5)(-9.8)(0.257)2 = 0.324 m

above the take-off point, and thus 5.324 m above the water.

(c) She hits the water at t1 = 1.3 seconds. So use the third and fourth *** equations to find vx(1.3) and vy(1.3). The answers are vx = 2.31 m/s and vy = -10.22 m/s. Thus v at this instant is 10.48 m/s at an angle of Arctan (-10.22/2.31) = -77.3 degrees (in the fourth quadrant).

P77: Make a coordinate system with (0, 0) at the release point. Our condition is that vy = 0 at the point where x = 9.0 and y = 8.0. Call the time when this happens t1. Then the four *** equations from P28 above, when tailored to this problem, give, at time t1:

x(t1) = (vo)xt1 = 9.0 m

y(t1) = (vo)yt1 - (4.9)t12 = 8.0

vx(t1) = (vo)x

vy(t1) = (vo)y - (9.8)t1 = 0

The third equation is not useful - it tells us that the x-component of velocity is not changing, but doesn't give us a way to find it. We have three other equations, and three unknowns: (vo)x, (vo)y, and t1.

We desire vx(t1) which is the same as (vo)x. This appears in the first equation, and we can solve it as soon as we know t1. The second and fourth equation contain t1 and (vo)y. SO: from fourth equation, (vo)y = (9.8)t1. Substitute this into the second equation and solve for t1 (obtaining t1 = 1.28 seconds). Then, finally, substitute this into first equation to obtain (vo)x = 7.04 m/s.

Comment: In this problem we encounter a set of three equations with three unknowns. I am unlikely to give a problem such as this on a Test, unless I simply ask you to set up the equations but not solve them.