# Answers/Solutions: Giancoli Chapter 3

Q7: The maximum value of V_{1} + V_{2} 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 V_{2}.

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** = 4.0**i**, then the magnitude A = 4.0 km, which is the same as A_{x}.

(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(v_{x}^{2} + v_{y}^{2})

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

P3: Magnitude is V = sqrt[(7.80^{2} + (-6.40)^{2}] = 10.1

Angle CCW from +x axis is where tan() = (-6.40/7.80). Thus tan() = -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

A_{x} = A cos(28) = (44.0)(0.883) = 38.8

A_{y} = A sin(28) = (44.0)(0.469) = 20.7

B_{x} = B cos(270) = (30.0)(0) = 0

B_{y} = 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** :

R_{x} = A_{x} - C_{x} = 38.8 - 0 = 38.8

R_{y} = A_{y} - C_{y} = 20.7 - (-30) = -50.7

Finally, R = sqrt[R_{x}^{2} + R_{y}^{2}] = sqrt[(38.8)^{2} + (-50.7)^{2})] = 63.8

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

tan() = R_{y}\R_{x} = -50.7/38.8 = -1.307

My calculator gives = Arctan(-1.307) = -52.6 degrees. Checking the components, we see that R_{x} is positive and R_{y} 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) = x
_{0}+ (v_{o})_{x}t + (1/2)a_{x}t^{2}

- x(t) = x

- y(t) = y
_{0}+ (v_{o})_{y}t + (1/2)a_{y}t^{2}

- y(t) = y

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

- v
_{x}(t) = (v_{o})_{x}+ a_{x}t

- v

- v
_{y}(t) = (v_{o})_{y}+ a_{y}t

- v

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, (v_{o})_{x} = +3.2 m/s, (v_{o})_{y} = 0. If we set the (x,y) origin on the ground directly under her leap, then x_{o} = 0 and y_{o} = +7.5 m. Finally, a_{y} = -g = -9.8 m/s^{2} and a_{x} = 0.

So, plugging in these values for x_{o}, y_{o}, (v_{o})_{x}, (v_{o})_{y}, a_{x}, and a_{y}, we have our mathematical model to describe the motion in this particular case:

x(t) = 3.2t

y(t) = 7.5 - 4.9t^{2}

We want x at the instant when y = 0. Call this time t_{1}. At this instant, the equations become:

x(t_{1}) = 3.2t_{1} = R

y(t_{1}) = 7.5 - 4.9(t_{1})^{2} = 0

From the second equation, t_{1} = (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 x_{o} = 0 and y_{o} = +1.0 m. We also have (v_{o})_{x} = v_{o}cos(_{o}) and (v_{o})_{y} = v_{o}sin(_{o}). Finally, we know that a_{x} = 0 and a_{y} = -g = -9.8 m/s_{2}.

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)t^{2} = (19.1)t - (4.9)t^{2}

The question (translated into mathspeak) is: what is the value of x when y = 13.0 m? Call the time at this point t_{1}. So at time t_{1}, we have

x(t_{1}) = (19.1)t_{1} = R

y(t_{1}) = (19.1)t_{1} - (4.9)t_{1}^{2} = 13.0

Two equations, two unknowns (R and t_{1}). We solve the second equation for t_{1}. The quadratic formula gives (as usual) two solutions: t_{1} = 0.88 sec and t_{1} = 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 t_{1} = 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 t_{1}. Then x_{o} = y_{o} = 0. Also a_{x} = 0 and a_{y} = -g = -9.8 m/s^{2}. Finally x(t_{1}) = +3 and y(t_{1}) = -5.

(a) In the first two equations, then, the unknowns are the x and y components of v_{o}. Plugging in the data at the instant t_{1} = 1.3 seconds:

+3 = (v_{o})_{x}(1.3)

-5 = (v_{o})_{y}(1.3) - (4.9)(1.3)^{2}.

Solutions are

(v_{o})_{x} = 2.31 m/s

(v_{o})_{y} = 2.52 m/s

So v_{o} = sqrt[(2.31^{2} + (2.52)^{2}] = 3.42 m/s at a angle of = 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 v_{y} = 0. Call this time t_{2}. We will solve the fourth *** equation in P28 for t_{2}, then use that value of time to find y(t_{2}) in the second *** equation. So:

0 = 2.52 - (9.8)t_{2}

which yields t_{2} = 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 t_{1} = 1.3 seconds. So use the third and fourth *** equations to find v_{x}(1.3) and v_{y}(1.3). The answers are v_{x} = 2.31 m/s and v_{y} = -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 v_{y} = 0 at the point where x = 9.0 and y = 8.0. Call the time when this happens t_{1}. Then the four *** equations from P28 above, when tailored to this problem, give, at time t_{1}:

x(t_{1}) = (v_{o})_{x}t_{1} = 9.0 m

y(t_{1}) = (v_{o})_{y}t_{1} - (4.9)t_{1}^{2} = 8.0

v_{x}(t_{1}) = (v_{o})_{x}

v_{y}(t_{1}) = (v_{o})_{y} - (9.8)t_{1} = 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: (v_{o})_{x}, (v_{o})_{y}, and t_{1}.

We desire v_{x}(t_{1}) which is the same as (v_{o})_{x}. This appears in the first equation, and we can solve it as soon as we know t_{1}. The second and fourth equation contain t_{1} and (v_{o})_{y}. SO: from fourth equation, (v_{o})_{y} = (9.8)t_{1}. Substitute this into the second equation and solve for t_{1} (obtaining t_{1} = 1.28 seconds). Then, finally, substitute this into first equation to obtain (v_{o})_{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.