19. A car speeds up from 12m/s to 25 m/s in 6.0 secs. What is its acceleration? How far didit travel in this time? Assume consistent acceleration.

You are watching: During what periods, if any, was the acceleration constant?

To begin, we deserve to notice that the initialvelocity of the automobile is not 0 m/s but is 12 m/s, and also then increases to 25m/s. The total time is 6 seconds.

-The change in velocity is justthe distinction in the 2 velocities. So the difference in between 12 m/s and25 m/s deserve to be calculated as:

25 m/s -12 m/s = 13 m/s

-This implies that the car provided anunknown acceleration to boost its velocity by 13 m/s.

-Acceleration is calculated by theformula: aav = Dv/Dt,so we deserve to put the numbers in and obtain this:

a= (13 m/s) / (6 sec)

Its acceleration is a= 2.2 m/s/s

-To uncover exactly how much it went in thistime, we deserve to use the formula v2=u2 + 2as, wbelow v is thelast velocity, u is the initial velocity, a is the acceleration and s is thedistance. Now put the numbers in:

(25 m/s)2= (12 m/s)2 + 2(2.2 m/s/s) s

481= 2(2.2) s

s= 1.1x102 m

Back

21. A light plane need to reach a rate of 30 m/s fortakeoff. How lengthy a runway is essential if the (constant) acceleration is 3.0m/s/s?

-The aircraft is beginning from remainder,and also therefore, its initial velocity, u, is 0 m/s. This helps out a greatdeal to note this. The final velocity, v, is 30 m/s. Theacceleration is 3.0 m/s/s.

-A formula we have the right to use is v2=u2 + 2as, wbelow v is the last velocity, u is the initial velocity, ais the acceleration and s is the distance. Now put the numbers in:

(30 m/s)2= (0 m/s)2 + 2(3.0 m/s/s) s

The initial velocity here is zero, so we can drop it from the equation.

900= 2(3.0 m/s/s) s

s= 1.5x102 m

Back

23. A car slows down from a speed of 25.0 m/s to restin 5.00 s. How much did it travel in this time?

-To begin, it is useful to figureout all that the question simply gave us. We recognize that the initial velocityis 25.0 m/s and that the final velocity is 0 m/s (it is at rest, meaning its notmoving). It takes a complete of 5 secs to do this. Also note that itis slowing down, so the acceleration is negative.

-We must first find what theacceleration is by utilizing this formula: aav = Dv/Dt, then put the numbers in and also deal with.

a= (25 m/s) / (5 sec)

a= - 5 m/s/s, bereason it is slowing down

-Now that we understand what theacceleration is, we have the right to deal with for the distance it takes to go from 25 m/s to 0m/s in 6 seconds. For this we deserve to use the formula: v2 = u2 + 2as,then put the numbers in and deal with.

(0 m/s)2= (25 m/s)2 + 2(-5 m/s/s) s

0= 625 + (-10) s

-625= (-10) s

s= 62.5 m

Back

51. The position of a rablittle bit alengthy a right tunnelas a function of time is plotted in Fig. 2-26. What is its instantaneousvelocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its averagevelocity (c) between t = 0 and t = 5.0 s, (d) in between t = 25.0 s and t = 30.0 s,and (e) between t = 40.0 s and t = 50.0 s?

*

-(a) Make a tangent line at t = 10s to find the instantaneous velocity by its slope. This is the less steepof the two blue lines above. This line seems to have a slope of rise/run =14 m/50 s = .28 m/s

-(b) Make a tangent line at t = 30s, and also uncover its slope (this is the velocity). This is the steeper of thetwo blue lines. This line rises 25 m between 17 and also 37 secs, so it hasa slope of about 25 m/20 s,which is about 1.25 or 1.2 m/s.

-(c) Median velocity is justcomplete displacement divided by full time. From 0 to 5 secs it seems togo from 0 to about 1.5 m (read from the graph), and the time is of course 5seconds, so the average velocity is 1.5 m/5 = .3 m/s

-(d) From 25 to 30 the rabbitdisareas from 8 m to 16 m (read from the graph), aacquire in 5 secs offering anaverage velocity of 8m/5 s = 1.6 m/s

-(e) From 40 to 50 seconds therablittle bit dislocations from 20 m to 10 m (read from the graph), currently in 10 secondsgiving an average velocity of -10m/10s = -1.0 m/s

Back

53. Figure 2-27 mirrors the velocity of a train as afunction of time. (a) At what time was its velocity greatest? (b) Duringwhat durations, if any, was the velocity constant? (c) Throughout what durations, ifany type of, was the acceleration constant? (d) When was the magnitude of theacceleration greatest?

*

-(a) Looking at the graph of v vs.time, we have the right to see that the best velocity was achieved approximately 50 s.

-(b) The velocity is constant whenthe velocity does not readjust, for if it does adjust, then tright here is anacceleration. The time period of 90-107 seconds has a consistent velocity.(it is additionally 0)

-(c) the acceleration is constantas soon as the velocity either rises or decreases at a constant price. Theselook prefer straight lines on a Velocity vs time graph. From 0to 20 s and also from 90 to 107 s (the acceleration was consistent, 0)

-(d) The magnitude of accelerationwas greatest once the slope of the velocity was steepest. This occurs ataround 75 s.

Back

57. Construct the v vs. t graph for the object whosedisplacement as a role of time is provided by Fig. 2-26.

*

-See the ago of the message book fora graph. To construct a graph of velocity vs. time from a distance vs.time graph, you end up taking the derivative (if you have actually already learnedderivatives) or you deserve to take the slope of the distance at particular points andgraph the different slopes, which ends up being the velocity.

The velocity graph is going to look prefer this:

interval Position Velocity
0-20 sec consistent slope constant V (horizontal line)
20-30 sec curve concave upwards positive acceleration - upward sloping straight line
30-37 sec curve concave downwards negative acceleration, downward sloping line that reaches zero
37-46 curve concave downwards negative acceleration, downward sloping line that proceeds down listed below zero (moving backwards)
46-50 curve, concave upwards positive acceleration, sloping line that goes up toward zero

Back

33. If a auto rolls gently (v=0) off a vertical cliff,exactly how long does it take to reach 90 km/h?

-First you have to adjust the 90km/h to m/s

90 km/h (1 hr / 3600 s)(1000 m / 1 km)

= 25 m/s

-You know that the initialvelocity (u) is 0, and that the final velocity (v) is 25 m/s.

-Using the formula v = u + at, wehave the right to discover the moment essential.

25 m/s = 0 + (9.8)t

t = 2.6 s

Back

35. Calculate (a) exactly how long it took King Kong to autumn straightdown from the peak of the Realm State Building (380 m high), and also (b) hisvelocity just before "landing"?

-We understand that the initial velocity(u) of King Kong is 0 m/s, and that the height of the structure is 380 m.

-(a) To discover the time, we have the right to usethe formula: s = ut + 1/2 at2

380 m = 0(t) + 1/2 (9.8)t2

t2 = 77.6161

t = 8.81 s

-(b) To uncover the last velocity(v), we have the right to usage the formula: v2 = u2 + 2as

v2 = (0)2 + 2(9.8)(380)

86.3 m/s

Back

37. A kangaroo jumps to a vertical elevation of 2.7m. How long was it in the air before returning to Earth?

-If the kangaroo jumps up, it mustautumn back to the ground. If it reaches a maximum height of 2.7 m, then werecognize that at this suggest, it is no longer going up, however is around to begin fallingearlier to the earth. At this allude, at 2.7 m over the earth, we understand thatthe last velocity (v) on the way up is zero, and also the initial velocity (u) onthe means dvery own is also zero.

-We can use the formula: s = ut +1/2 at2. Now put in the indevelopment we recognize.

2.7 = 0(t) + 1/2(9.8)t2

2.7 = 4.9t2

t = .74 s

-But since this is only on the waydvery own, the time should be doubcaused encompass the time it was going up.

.74 x 2 = 1.48 which is about 1.5 s

Back

41. A helicopter is ascfinishing vertically via a speedof 5.5 m/s. At a height of 105 m above the Earth, a package is droppedfrom a home window. How much time does it take for the package to reach theground?

-The package and the helicopterare both 105 m over the earth, but are both going up at 5.5 m/s. First weneed to discover out the amount of time the package spends going up, before it beginsto go down to the earth. The last velocity (v) for when the package isgoing up is zero, bereason at this moment, the package isn"t going up anyeven more,however is falling toward the ground. We have the right to use the formula: v = u + at

0 = 5.5 + (-9.8)t

-5.5 = (-9.8)t

t = .56 s

-Now we understand just how long the packagespends going up. But exactly how high does the package go above the 105m? Remember that the initial velocity (u) is zero. For this wehave the right to usage the formula: v2 = u2 + 2as.

02 = (5.5)2 +2(-9.8)s

s = 1.54 m

-Add this to the 105 m, and we get106 .54 m.

-How long does it take for thepackage to drop from this height, 106.54 m? We have the right to usage the formula: s = ut+ 1/2 at2

106.54 = (0)t + 1/2(9.8)t2

t = 4.66 s

-Remember that the package wasgoing up initially and took .56 s to perform that. Add 4.66 s and also .56 s to obtain 5.22s.

Back

47. A rock is dropped from a sea cliff and the soundof it striking the sea is hears 3.4 seconds later on. If the speed of soundis 340 m/s, exactly how high is the cliff?

-We can use the formula s = ut +1/2 at2, but the initial velocity (u) is zero, so we are left through, s = 1/2 at2.Replacing h through s, and also addressing for time offers us tf (the moment tofall) = (2h)/g)^1/2

-The time it takes for the soundto be heard up at the optimal of the cliff deserve to be discovered by the formula v = s/t or t= s/v, which is ts = h /(340 m/s).

-The full time is 3.4 s, so ts+ tf = 3.4 s

-Substitute for ts andtf. This provides us:

((h /340) + (2h /g)^1/2 = 3.4)

-This is the exact same as:

(1/340) ((h)^1/2)2 + (2/g)^1/2(h)^1/2 - 3.4 = 0

-This is quadratic currently, wbelow x =h^1/2, so deal with it making use of the quadratic formula (in your book on page1042).

h = 52 m

Back

49. A stone is thrown vertically upwardthrough a speed of 12.0 m/s from the edge of a cliff 75.0 m high (Fig. 2-32).(a) How much later on does it reach the bottom of the cliff? (b) What is itsspeed just before hitting? (c) What total distance did it travel?

-(a) If it is thrownupward, we should uncover the moment it takes to soptimal going upward, and also reach a finalvelocity (v) of zero. Using the formula v = u + at, discover the time in theair.

0 = 12 + (-9.8)t

t = 1.23 s

-Now find the heightit went in the time of that 1.22 s making use of the formula v2 = u2 +2as.

02 = (12)2 + 2(-9.8)s

s = 7.35 m

-Add 7.35 m to theelevation of the cliff, 75 m, to get 82.35 m. Now discover the time it takes toautumn this brand-new elevation making use of the formula s = ut+ 1/2 at2

82.35 m = (0)t + 1/2(-9.8)t2

t = 4.1 s

-Add 4.1 s to thetime the stone is going upward, which is 1.23 s, and you get 5.33 s.

-(b) The initialvelocity (u) is 0 m/s and the elevation is 82.35 m. Using the formula v2= u2 + 2as, uncover the last velocity.

v2 = (0)2 + 2(-9.8)(82.35)

v = - 40.2 m/s (negative bereason it is going down)

-(c) The totaldistance it traveled would certainly be the distance it traveled going up, 7.35 m, and thecomplete distance it visited the bottom of the cliff, 82.35 m. So addingthese together, we gain, 89.7 m.

Back

25. A car traveling 45 km/h slows downat a continuous .50 m/s/s simply by "letting up on the gas".Calculate (a) the distance the automobile shores before it stops, (b) the time it takesto soptimal and also (c) the distance it travels during the first and fifth secs.

-Change 45 km/h intom/s

45 km/h ((1 hr)/(3600 s)) ((1000 m)/(1 km)) = 12.5 m/s

-Remember thatslowing dvery own is an unfavorable acceleration. So, a = - .5 m/s/s

-(a) The initialvelocity (u) is 12.5 m/s, the acceleration is -.5 m/s/s, and also the final velocity(v) is 0 m/s, because we desire it to speak. Using the formula v2= u2 + 2as, calculate the distance it takes to stop.

02 = (12.5)2 +2(-.5)s

s = 156.25 m or 1.6 x 102 m

-(b) To uncover the timeit takes to stop, put the indevelopment in to the formula: v = u + at.

0 = 12.5 + (-.5)t

t = 25 s

-(c) The distance thevehicle traveled in the time of the initially second have the right to be found by making use of the formula s = ut +1/2 at2, and substituting 1 s in for t.

s = (12.5)(1) + (.5)(-.5)(1)2

s = 12.25 m or 12 m

-Throughout the fifthsecond, you have to realize that the distance traveled during this time is thedistance between the fifth and 6th seconds. So to uncover the distancethroughout only the fifth second, you take the distance traveled up to the sixthsecond, and also subtract the distance traveled up to the fifth second.

s = (12.5)(6) + (.5)(-.5)(6)2s = 66 m

s = (12.5)(5) + (.5)(-.5)(5)2s = 56.25 m

Now take the distance up to the 6th second and also subtract the distance as much as thefifth.

66 m - 56.25 m = 9.75 m or 10 m

Back

27. Determine the stopping ranges foran automobile with an initial rate of 90 km/h and a huguy reactivity time of1.0s: (a) for an acceleration a= - 4.0 m/s/s; (b) for an acceleration a= - 8m/s/s.

-Change 90 km/h intom/s.

90 km/h ((1 hr)/(3600 s)) ((1000 m)/(1 km)) = 25 m/s

-(a) First, uncover thedistance it takes to sheight via an acceleration of - 4.0 m/s/s. Use theformula: v2 = u2 + 2as

02 = 252 + 2(-4.0)s

s = 78.125 m

-But remember that ittakes one second for huguy reaction, so before the driver starts slowing dvery own,the auto is traveling at 25 m/s for one second. The car travels a distanceof 25 m.

-Add 25 m to the78.125 m, and also you get 103 m.

-(b) Do the same asvia a, however for an acceleration of -8.0 m/s/s.

See more: This Is What You Came For Piano : This Is What You Came For By Calvin Harris

02 = 252 + 2(-8.0)s

s = 39.06 m

-Add 39.06 m to the25 m the car travels throughout the one second of huguy reaction to acquire 64 m.