# Answers/Solutions: Giancoli Chapter 2

Q2: A specified velocity has both a definite speed and a specific direction. If either one changes, the velocity will not be the same as before. So: NO - any change in speed will change the velocity.

YES: since a specific velocity has both a given speed AND a specific direction, we can change the velocity by changing the DIRECTION while keeping the speed constant.

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Q6: Absolutely. If you are moving northward but slowing your speed, your acceleration (rate of change of velocity) would be in a southward direction.

For example: Call north positive. Assume that at time t_{1} = 0 your speed is +10 m/s and at time t_{2} = 2 seconds your speed is +4 m/s. Then your (average) acceleration would be (v_{2} - v_{1})/(t_{2} - t_{1}) = (4 - 10)/(2 - 0) = -6/2 = -3 m/s. The minus indicates southward.

A second example: Any time an object is tossed upward, then during its flight upward (before it momentarily stops and begins its descent, it has velocity up but acceleration down (at 9.8 m/s^{2}).

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Q14: Yes. Consider an object tossed in the air. Its acceleration is downward at 9.8 m/s^{2} during its travel up and back down. At the instant when it reaches the "peak" of its path, its velocity is zero. Thus, at that moment, its velocity is zero while its acceleration is 9.8 m/s^{2} downward.

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Q15: Of course. An object moving at any **constant** velocity (whethere zero or not) has no acceleration. So, let the speed be an unchanging 5 m/s with motion in a steady northward direction. Then v = 5 m/s north, and the acceleration = 0.

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P6: Divide the trip into two parts. In part I you travel a distance of 130 km at a speed of 95 km/hr. The time of travel for part I will be (130 km)/(95 km/hr) = 1.37 hours. Then, since the total time of travel (parts I and II combined) is 3.33 hours, the time for part II must be 3.33-1.37 = 1.96 hours. Since the speed for part II is 65 km/hr, the distance for part II must be (65 km/hr)x(1.96 hr) = 127 km. We now have ALL the information for each part.

The total distance traveled is 130 + 127 = 257 km.

The average speed is (257 km)/(3.33 hr) = 77.2 km/hr

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P16: Examine Example 3 on page 24 closely. This problem uses exactly the same logic. The answers are (with x in meters, v in meters/sec):

(a) x(1) = -0.5; x(2) = -0.8; x(3) = +1.1

(b) <v> = +0.8

(c) v(2) = +0.8; v(3) = +3.0

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P27: Examine Example 7 on page 27 closely. In this problem you need to differentiate the function for x(t) twice: the first derivative yields the speed v(t) and the second derivative then gives the acceleration a(t). The answer is +17.0 m/s^{2} (so the acceleration a is not changing: v increases by 17.0 m/s each second).

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P42: We have the speeds at two times, so a simple division gives the acceleration: a = v/t = (162 - 65)/(10 - 0) = 9.7 m/s^{2}.

Knowing the acceleration, setting the initial position x_{o} = x(0) to zero, and using v_{o} = v(0) = 65 m/s, we can solve for x(t) at any time t using equation 2-12(b). Thus:

x(2) = 0 + (65)(2) + (0.5)(9.7)(2)^{2} = 149.5 m

x(6) = 0 + (65)(6) + (0.5)(9.7)(6)^{2} = 566.4 m

so x = 417 m

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P54: From the instant the jumping player leaves the court surface until she again touches it, her motion is governed by Eqns. 2-12. If "upward" is chosen as the positive direction, then a = = -g = -9.8 m/s_{2}. We can let the starting position (y_{o}) be zero. (See the short paragraph just above Exercise H on page 35.)

In this problem we are given that v = 0 when y = 1.20 m. We can use Eqn. 2-12(a) to solve for the time (label it t_{1}) when v is zero: the answer is t_{1} = v_{o}/g. Now substitute this into 2-12(b) and you find that y(t_{1}) - which should be the maximum height - is v_{0}^{2}/2g. Since we know that this is 1.20 m, we can immediately solve for v_{o} finding that it equals 4.85 m/s.

Knowing the initial speed, we can use Eqn. 2-12(b). Let t_{2} be the time when the player comes down, then y(t_{2}) = 0 and solve for t_{2} using the quadratic formula. One solution will be t_{2} = 0, which tells us that y = 0 at t = 0, which we already knew. Obviously we want the OTHER solution, which is t_{2} = 2v_{o}/g = 0.99 sec.

(What about the bogus solution? The equations don't know *when* the motion started. As far as the equations are concerned, the player has ALWAYS been in free-fall, and always will be. But our clock starts when she jumps (y_{o} = 0). For negative values of time, the equations think that the player was below court level, on her way up, and for very large times, the equations think that she is again below the court, accelerating downward forever! But the PHYSICS of the situation tells us that the DOMAIN of the functions given by Eqns. 2-12 is - in this case - only for the time interval between t = 0 and t = t_{2}. The POINT: when you solve a quadratic such as Eqn. 2-12(b) for time you will always get two solutions. You have to use physical considerations to decide which one is the answer you seek.)

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P81: The motion of the stone is governed by Eqns. 2-12. Take upward to be the positive direction. We are given that v_{o} = +12.5 m/s, and we know a = -g = -9.8 m/s^{2}. Set y = 0 at the top of the cliff (as shown in Figure 2-49).

(a) If we label the time when the diver hits the water as t_{2} (reserving t_{1} to label the "highest point" time), then we simply use y(t_{2}) = -75 in Eqn. 2-12(b) and solve for t_{2}. (Again, we get two answers. Clearly we want the positive value. Can you understand why, and what the negative value means?) Our answer is +5.39 sec.

(b) We want v(t_{2}) which, now that we know t_{2}, we can get directly from Eqn. 2-12(a). Take care with signs: v_{o} is positive, a is negative. The answer is -40.3 m/s (the minus indicating downward).

(c) NOW we must find out how far ABOVE the cliff the diver rises before starting back down. This requires finding t_{1}, which we again do by setting v(t_{1}) = 0 in Eqn. 2-12(a), and then using this time in Eqn. 2-12(b) to find y(t_{1}). The results are:

the time from launch to highest point is 1.28 sec

the distance of highest point above launch point is y(1.28) = 7.97 m

Finally, The total distance traveled is 2y(t_{1}) + 75 (i.e., distance from cliff to peak + distance from peak back to cliff + distance from cliff to water). Using our value for y_{1}, we get the total distance traveled to water is 2(7.97) + 75.0 = 90.9 m