PH 2213 : Chapter 02 Homework Solutions Problem 2.6 : You are driving home from school steadily at 90 km/hr for 130 km. It then begins to rain and you slow to 50 km/hr. You arrive home after driving 3 hours and 20 minutes. (a) How far is your hometown from school? We have two segments here where we re driving at different speeds. During the first part of the trip, we travelled 130 km at a constant speed of 90 km/hr. How long did this segment take? v avg = x/ t so t = x/v avg = (130 km)/(90 km/hr) = 1.4444.. hour. During the second part of the trip, we travelled some unknown distance at a speed of 50 km/hr. We do know the entire trip took 3 hours and 20 minutes, or converting to hours that would be 3.333... hours. The first part took 1.4444.. hours so this second segment of the trip must have taken 3.3333.. 1.4444... = 1.8888.. hour. During the second part of the trip then we travelled at 50 km/hr for 1.8888... hour so x = v avg t = (50 km/hr)(1.8888... hr) = 94.444.. km. Adding the distances covered in each of the two segments, the total distance travelled was: (130 km)+ (94.444... km) = 224.44.. km. (The online system asked for the answer with just two significant figures, so you could round this so 220 km.) (b) What was your average speed? Over the entire trip, we covered a distance of 224.444... km in 3.3333 hour so the average speed over the entire trip would be v avg = x/ t = (224.444... km)/(3.333... hr) = 67.3333.. km/hr. (Rounded to two significant figures, this would be 67 km/hr.)
Problem 2.16 : The position of a ball rolling in a straight line is given by x(t) = 2.0 3.6t + 1.1t 2 where x is in meters and t in seconds. (a) Determine the position of the ball at t = 1 s, t = 2 and t = 3. Here we just evaluate x(t) at the given values of t, yielding positions of 0.50 m, 0.80 m, and +1.10 m respectively. (b) What is the average velocity over the interval t = 1 to t = 3? v avg = x/ t = x(3) x(1) 3 1 = (1.10) ( 0.5) 3 1 = +0.8 m/s. We have some other options here. If we differentiate the position, we have v(t) = dx/dt = 3.6 + 2.2t. Differentiating again, a(t) = dv/dt = 2.2 m/s 2 which is constant. Since we have a constant acceleration here, we have other options to solve this, or to check the previous result. If a is constant, the average velocity is equal to the average of the instananeous velocities at the two points. Evaluating v(1) = 1.4 m/s and v(3) = +3.0 m/s yielding v avg = ( 1.4)+(3.0) = 2 +0.8 m/s. With constant acceleration, the average velocity is also equal to v(t) evaluated at the timemidpoint, so since we ve looking at t = 1 to t = 3 the midpoint will be at t = (1 + 3)/2 = 2 sec. Evaluating v(t) at t = 2 we get v avg = v(2) = 3.6 + 2.2(2) = +0.8 m/s. (Note that both of those last bits required constant acceleration, which we did have in this case...) (c) What is its instantaneous velocity at t = 2 and t = 3? Above, we found v = dx/dt = 3.6 + 2.2t and evaluating this expression at the given times: v(2) = 3.6 + 2.2(2) = +0.8 m/s and v(3) = 3.6 + 2.2(3) = +3.0 m/s.
Problem 2.40 : A car travelling at 105 km/hr strikes a tree. The front end of the car compresses and the driver comes to rest after travelling 0.80 m. What was the magnitude of the average acceleration of the driver during the collision? Express the answer in terms of g s, where 1.00 g = 9.80 m/s 2. The person goes from a velocity of 105 km/hr to 0 km/hr over the given distance, so we can use our v 2 equation to determine the acceleration: v 2 = v 2 o + 2a x. Converting the initial speed to standard metric units: 105 km 1 hr 1000 m 1 km 1 hr 3600 s Using a coordinate system pointing in the direction of the motion of the car: = 29.167 m/s. (0) 2 = (29.167) 2 + (2)(a)(0.80) from which a = 531.7 m/s 2. They re just asking for magnitude here, which would be 531.7 m/s 2. They also want us to convert this into units of g s, so: 531.7 m/s 2 1 g 9.80 m/s 2 = 54.3 g s.
Problem 2.58 : A rocket rises vertically, from rest, with an acceleration of 5.0 m/s 2 until it runs out of fuel at an altitude of 990 m. OVERVIEW: looking over the questions, we have a rocket that s accelerating upward for a while, then the rocket engine cuts off. For the remainder of the flight, the rocket will be undergoing a different acceleration: that of g downward. All our equations of motion require constant acceleration, so we ll need to focus on these two acceleration intervals separately, and fortunately the questions are asked in a way that leads us through the steps involved. (a) What is the velocity of the rocket when it runs out of fuel? For this part of the flight, the rocket has a constant acceleration (upward) of 5.0 m/s 2. Let s use a coordinate system with +Y pointing vertically upward, with Y = 0 at ground level. We know the initial velocity (0 since the rocket started from rest), we know the height it reaches and we know the acceleration during this interval so we can use: v 2 = v 2 o + 2a y to find the velocity at this point: v 2 = (0) 2 + (2)(5 m/s 2 )(990 m) from which v = 9900 = ±99.4987.. m/s. At this point the rocket is moving upward, so in our chosen coordinate system it s velocity would be v = +99.4987.. m/s. (The homework system wanted this rounded to two significant figures, so we could enter v = 99 m/s.) (b) How long does it take to reach this point? During this part of the flight, we know the initial velocity (zero), the final velocity (99.4987.. m/s) and the acceleration (5 m/s 2 ) and we re looking for the time, so let s use v = v o + at : 99.4987 = 0 + 5t from which t = 19.8997.. sec. (Rounded to two significant figures, this would be just 20 sec.) (c) What maximum height does the rocket reach? When the rocket motor cuts out, it s now in free fall where it undergoes a downward acceleration of 9.8 m/s 2. It s still moving upward but it s slowing down now, eventually reaching a point where it s velocity becomes zero and it no longer moves upward; instead it starts falling back towards the ground. So at the point of maximum height, that s where its velocity has become zero. Still using our original coordinate system, v 2 = v 2 o + 2a y. Over this interval, the rocket is starting at y = 990 m and has an initial velocity of +99.4987 m/s. When it reaches it s maximum altitude, it s velocity is zero. Over this part of the trip, the acceleration is a = 9.8 m/s 2 so v 2 = v 2 o + 2a y becomes: (0) 2 = (99.4987 m/s) 2 + (2)( 9.8) y from which y = 505.1 m. That s the change in it s Y coordinate though from where it started during this interval, so that puts it 505.1 m above where this interval started, which was at 990 m above the ground, so the total height at this point is 990 + 505.1 = 1495.1 m. (Rounded to two significant figures, the homework system accepted 1500 m for the maximum height.) Continued...
(d) How much time (total) does it take to reach maximum altitude? We have two different acceleration regimes here. During the first part of the flight, the rocket was accelerating upward at 5 m/s 2 for 19.8997 s, reaching an altitude of 990 m. Then it switched to an interval where the rocket started slowing down due to gravity. During that interval, the velocity dropped from 99.4987 m/s to 0 m/s with an acceleration of a = 9.8 m/s 2 so applying v = v o + at to this interval we have (0) = 99.4987 9.8t or t = 10.153 sec. That s the time just for this interval though, so the total time to reach this point is 19.8997 + 10.153 = 30.05 sec. (e) With what velocity does it strike the Earth? Let s look at the interval where the rocket is going from it s maximum height back to the ground. At the start, it s located 1495.1 m above the ground and it s accelerating with a = 9.8 m/s 2 until it hits the ground. v 2 = v 2 o + 2a y where y = y final y initial = 0 1495.1 = 1495.1 so v 2 = v 2 o + 2a y implies that: v 2 = (0) 2 + (2)( 9.8)( 1495.1) = 29303.96 from which v = ±171.18 m/s. The rocket is travelling downward in our coordinate system, so technically v = 171.2 m/s (or rounded to two significant figures, v = 170 m/s). (f) How long (total) is it in the air? From part (d), we already know it took 30.05 sec for the rocket to get from the ground up to it s maximum height. How about the time it takes to get from that point back to the ground? Over that interval, we know the rocket started with a velocity of 0 (up at it s maximum height), and ended up travelling at v = 171.2 m/s when it hits the ground, and during that interval it has an acceleration of a = 9.8 m/s 2 so: v = v o + at becomes 171.2 = 0 9.8t or t = 17.467 sec. The total trip then from launch to landing took 30.05 + 17.467 = 47.52 s or rounded to two significant figures, that would be 48 sec. NOTE: the next page has some graphs showing the acceleration, velocity, and height of the rocket from launch to landing.
Top figure : acceleration the rocket is accelerating upward at 5.0 m/s 2 for a while (which we determined to be just under 20 seconds). The rocket then cuts out and for the remainder of the rocket s flight, gravity is accelerating it downward at 9.8 m/s 2. The figures are all labeled such that: a is the initial location on the ground, at rest, accelerating upward. b is the point where the rocket stops firing. c is the point where the rocket reaches its maximum altitude. d is the point where the rocket has returned to the ground. Middle figure : velocity From (a) to (b), the rocket is accelerating upward at 5 m/s 2 so it s velocity keeps increasing up to point (b). At point (b), the rocket has cut off. It picked up a considerable velocity by this point, so it s slowing down but the velocity is still positive (for a while anyway) so the rocket is continuing to move upward. At point (c), it s vertical velocity has dropped to zero and now becomes negative. This represents the point where the rocket has reached it s maximum height and now starts falling back towards the ground. At point (d), it collides with the ground. Bottom figure : position From (a) to (b), the rocket is accelerating upward, so its position is described by an upward curving parabola. After the rocket cuts out at (b), for the remainder of the trip we have a negative acceleration, so the position function is described by a downward curving parabola.
Problem 2.62 : Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of 1.5 m above the ground. When you quickly turn off the nozzle, you hear the water striking the ground next to you for another 2.0 s. What is the water speed as it leaves the nozzle? If we focus on that last drop of water to leave the hose, it s moving upward at some unknown velocity v o, continues to travel upward slowing down under the influence of gravity, eventually stopping and falling back down, ultimately hitting the ground 2.0 s after it left the hose. Using a Y axis starting on the ground level and pointing upward, what do we know? We know the initial and final y values, the acceleration, and the time, but we do not know the initial velocity. Looking through our equations of motion, we can use y = y o + v o t + 1 2 at2 to find this initial velocity. With the initial position being the point where the drop exits the hose and the final position being the point where the drop hits the ground: (0) = (1.5 m) + (v o )(2.0 s) + 1 2 ( 9.8 m/s2 )(2.0 s) 2 or 0 = 1.5 + 2.0v o 19.6. Rearranging: 19.6 1.5 = 2.0v o or v o = 9.05 m/s ALTERNATIVE: The average velocity over an interval is defined as v avg = y/ t so from 0 to 2 seconds, we have y = y final y initial = 0 1.5 so v avg = 0.75 m/s over this 2 second interval. Since we have constant acceleration here, the average velocity is the value of v at the midpoint of the interval. v = v o + at so v avg = v o + 1 2 at and here: 0.75 = v o + (0.5)( 9.8)(2) or 0.75 = v o 9.8 or v o = 9.8 0.75 = 9.05 m/s. That was a bit tricky but at least provided a use for our average velocity equations...
Problem 2.64 : A ball is dropped from the top of a 55 m high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff at a speed of 26 m/s. The stone and ball collide part way up. (a) How far above the base of the cliff does this happen? The two objects are launched at the same time, so let s use a common coordinate system with +Y pointing vertically upward with the origin of coordinates down on the ground at the bottom of the cliff. BALL: this object is dropped (released at rest) at the top of the cliff, and from there on it s accelerating downward at 9.8 m/s 2. The generic equation of motion for it s height: y = y o + v o t + 1 2 at2 becomes: y ball = 55 + 0 4.9t 2. STONE: this object is launched vertically upward from the base of the cliff at 26 m/s, after which it s in free fall and is also accelerating downward at 9.8 m/s 2 so the generic equation of motion for its height: y = y o + v o t + 1 2 at2 becomes: y stone = 0 + 26t 4.9t 2. When they run into each other, they ll both have the same y coordinate, so setting these equations equal to each other: y ball = y stone yields 55 4.9t 2 = 26t 4.9t 2 and cancelling the common term on each side we have 55 = 26t or t = 55/26 = 2.11536.. sec. That gives us the time when the two objects meet. Calculating their positions: y ball = 55 4.9t 2 = 55 (4.9)(2.11536..) 2 = 33.07 m. y stone = 0 + 26t 4.9t 2 = (26)(2.11526..) (4.9)(2.11526..) 2 = 33.07 m (did both to make sure they came out the same...) ADDENDUM: what velocity does each object have when they meet? In general, v = v o + at so for the ball dropped from the top of the cliff we have v ball = 0 9.8t and evaluating that at t = 2.11536.. sec we have v ball = 20.73 m/s. For the stone launched vertically, we have v stone = 26 9.8t and evaluating that at t = 2.11536.. sec we find v stone = +5.269.. m/s. (That s positive, which means when they collide, the stone is still moving upward.) These are quite different in magnitude and in direction.