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(a) The rider’s displacement is Δx=xfx0=−1kmΔx=xfx0=−1km. (The displacement is negative because we take east to be positive and west to be negative.) (b) The distance traveled is 3 km + 2 km = 5 km. (c) The magnitude of the displacement is 1 km.


(a) Taking the derivative of x(t) gives v(t) = −6t m/s. (b) No, because time can never be negative. (c) The velocity is v(1.0 s) = −6 m/s and the speed is |v(1.0s)|=6m/s|v(1.0s)|=6m/s.


Inserting the knowns, we have


If we take east to be positive, then the airplane has negative acceleration because it is accelerating toward the west. It is also acceleration opposite to the motion; its acceleration is opposite in direction to its velocity.


To answer this, choose an equation that allows us to solve for time t, given only a , v0 , and v:
Rearrange to solve for t:




It takes 2.47 s to hit the water. The quantity distance traveled increases faster.

  1. The velocity function is the integral of the acceleration function plus a constant of integration. By Equation 3.18,
    Since v(0) = 0, we have C1 = 0; so,
  2. By Equation 3.19,
    Since x(0) = 0, we have C2 = 0, and
  3. The velocity can be written as v(t) = 5t(1 – t), which equals zero at t = 0, and t = 1 s.

Conceptual Questions


You drive your car into town and return to drive past your house to a friend’s house.


If the bacteria are moving back and forth, then the displacements are canceling each other and the final displacement is small.


Distance traveled


Average speed is the total distance traveled divided by the elapsed time. If you go for a walk, leaving and returning to your home, your average speed is a positive number. Since Average velocity = Displacement/Elapsed time, your average velocity is zero.


Average speed. They are the same if the car doesn’t reverse direction.


No, in one dimension constant speed requires zero acceleration.


A ball is thrown into the air and its velocity is zero at the apex of the throw, but acceleration is not zero.


Plus, minus


If the acceleration, time, and displacement are the knowns, and the initial and final velocities are the unknowns, then two kinematic equations must be solved simultaneously. Also if the final velocity, time, and displacement are the knowns then two kinematic equations must be solved for the initial velocity and acceleration.


a. at the top of its trajectory; b. yes, at the top of its trajectory; c. yes


Earth v=v0gt=gtv=v0gt=gt; Moon v=g6tv=vgt=g6tt=6tv=g6tv=vgt=g6tt=6t; Earth y=12gt2y=12gt2 Moon y=12g6(6t)2=12g6t2=−6(12gt2)=−6yy=12g6(6t)2=12g6t2=−6(12gt2)=−6y



a. x1=(−2.0km)i^x1=(−2.0km)i^, x2=(5.0km)i^x2=(5.0km)i^; b. 7.0 km east


a. t=2.0t=2.0 s; b. x(6.0)x(3.0)=−8.0(−2.0)=−6.0mx(6.0)x(3.0)=−8.0(−2.0)=−6.0m


a. 150.0 s, v=902m/sv=902m/s; b. 163% the speed of sound at sea level or about Mach 2.

Graph shows velocity in meters per second plotted as a function of time at seconds. Velocity starts as 10 meters per second, decreases to -30 at 0.4 seconds; increases to -10 meters at 0.6 seconds, increases to 5 at 1 second, increases to 15 at 1.6 seconds.
Graph shows position plotted versus time. It starts at the origin, increases reaching maximum, and then decreases close to zero.

a. v(t)=(104t)m/sv(t)=(104t)m/s; v(2 s) = 2 m/s, v(3 s) = −2 m/s; b. |v(2s)|=2m/s,|v(3s)|=2m/s|v(2s)|=2m/s,|v(3s)|=2m/s; (c) v=0m/sv=0m/s


a = 4.29 m/s 2 a = 4.29 m/s 2

Graph shows acceleration in meters per second squared plotted versus time in seconds. Acceleration is 0.3 meters per second squared between 0 and 20 seconds, -0.1 meters per second squared between 20 and 50 seconds, zero between 50 and 70 seconds, -0.6 between 90 and 100 seconds.

a = 11.1 g a = 11.1 g


150 m


a. 525 m;
b. v=180m/sv=180m/s



Graph is a plot of acceleration a as a function of time t. Graph is non-linear with acceleration being positive at the beginning, negative at the end, and crossing x axis between points d and e and at points e and h.

b. The acceleration has the greatest positive value at tata
c. The acceleration is zero at teandthteandth
d. The acceleration is negative at ti,tj,tk,tlti,tj,tk,tl


a. a=−1.3m/s2a=−1.3m/s2;
b. v0=18m/sv0=18m/s;
c. t=13.8st=13.8s


v = 502.20 m/s v = 502.20 m/s



Figure shows object with speed equal to 0 meters per second and acceleration equal to 2.4 meters per second squared at zero point. When time is equal to 12 seconds, acceleration remains equal to 2.4 meters per second. Speed and position of the object are unknown.

b. Knowns: a=2.40m/s2,t=12.0s,v0=0m/sa=2.40m/s2,t=12.0s,v0=0m/s, and x0=0mx0=0m;
c. x=x0+v0t+12at2=12at2=2.40m/s2(12.0s)2=172.80mx=x0+v0t+12at2=12at2=2.40m/s2(12.0s)2=172.80m, the answer seems reasonable at about 172.8 m; d. v=28.8m/sv=28.8m/s



Figure shows object with zero speed and unknown acceleration at the beginning. After unknown time, object reaches speed of 30 centimeters per second and is at distance of 1.8 centimeters from the starting point. Acceleration of the object at this point is unknown.

b. Knowns: v=30.0cm/s,x=1.80cmv=30.0cm/s,x=1.80cm;
c. a=250cm/s2,t=0.12sa=250cm/s2,t=0.12s;
d. yes


a. 6.87 m/s2; b. x=52.26mx=52.26m


a. a=8450m/s2a=8450m/s2;
b. t=0.0077st=0.0077s


a. a=9.18g;a=9.18g;
b. t=6.67×10−3st=6.67×10−3s;
c. a=40.0m/s2a=4.08ga=40.0m/s2a=4.08g


Knowns: x=3m,v=0m/s,v0=54m/sx=3m,v=0m/s,v0=54m/s. We want a, so we can use this equation: a=486m/s2a=486m/s2.


a. a=32.58m/s2a=32.58m/s2;
b. v=161.85m/sv=161.85m/s;
c. v>vmaxv>vmax, because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears and would have a greater acceleration in first gear than second gear than third gear, and so on. The acceleration would be greatest at the beginning, so it would not be accelerating at 32.6m/s232.6m/s2 during the last few meters, but substantially less, and the final velocity would be less than 162m/s162m/s.


a. y=−8.23mv1=18.9m/sy=−8.23mv1=18.9m/s;
b. y=−18.9mv2=−23.8m/sy=−18.9mv2=−23.8m/s;
c. y=−32.0mv3=28.7m/sy=−32.0mv3=28.7m/s;
d. y=−47.6mv4=33.6m/sy=−47.6mv4=33.6m/s;
e. y=−65.6mv5=38.5m/sy=−65.6mv5=38.5m/s


a. Knowns: a=9.8m/s2v0=−1.4m/st=1.8sy0=0ma=9.8m/s2v0=−1.4m/st=1.8sy0=0m;
b. y=y0+v0t12gt2y=v0t12gt=−1.4m/s(1.8sec)12(9.8)(1.8s)2=−18.4my=y0+v0t12gt2y=v0t12gt=−1.4m/s(1.8sec)12(9.8)(1.8s)2=−18.4m and the origin is at the rescuers, who are 18.4 m above the water.


a. v2=v022g(yy0)y0=0v=0y=v022g=(4.0m/s)22(9.80)=0.82mv2=v022g(yy0)y0=0v=0y=v022g=(4.0m/s)22(9.80)=0.82m; b. to the apex v=0.41sv=0.41s times 2 to the board = 0.82 s from the board to the water y=y0+v0t12gt2y=−1.80my0=0v0=4.0m/sy=y0+v0t12gt2y=−1.80my0=0v0=4.0m/s−1.8=4.0t4.9t24.9t24.0t1.80=0−1.8=4.0t4.9t24.9t24.0t1.80=0, solution to quadratic equation gives 1.13 s; c. v2=v022g(yy0)y0=0v0=4.0m/sy=−1.80mv=7.16m/sv2=v022g(yy0)y0=0v0=4.0m/sy=−1.80mv=7.16m/s


Time to the apex: t=1.12st=1.12s times 2 equals 2.24 s to a height of 2.20 m. To 1.80 m in height is an additional 0.40 m. y=y0+v0t12gt2y=−0.40my0=0v0=−11.0m/sy=y0+v0t12gt2y=−0.40my0=0v0=−11.0m/s0.40=−11.0t4.9t2or4.9t2+11.0t0.40=0y=y0+v0t12gt2y=−0.40my0=0v0=−11.0m/sy=y0+v0t12gt2y=−0.40my0=0v0=−11.0m/s0.40=−11.0t4.9t2or4.9t2+11.0t0.40=0.
Take the positive root, so the time to go the additional 0.4 m is 0.04 s. Total time is 2.24s+0.04s=2.28s2.24s+0.04s=2.28s.


a. v2=v022g(yy0)y0=0v=0y=2.50mv02=2gyv0=2(9.80)(2.50)=7.0m/sv2=v022g(yy0)y0=0v=0y=2.50mv02=2gyv0=2(9.80)(2.50)=7.0m/s; b. t=0.72st=0.72s times 2 gives 1.44 s in the air


a. v=70.0m/sv=70.0m/s; b. time heard after rock begins to fall: 0.75 s, time to reach the ground: 6.09 s


a. A=m/s2B=m/s5/2A=m/s2B=m/s5/2;
b. v(t)=a(t)dt+C1=(ABt1/2)dt+C1=At23Bt3/2+C1v(0)=0=C1sov(t0)=At023Bt03/2v(t)=a(t)dt+C1=(ABt1/2)dt+C1=At23Bt3/2+C1v(0)=0=C1sov(t0)=At023Bt03/2;
c. x(t)=v(t)dt+C2=(At23Bt3/2)dt+C2=12At2415Bt5/2+C2x(0)=0=C2sox(t0)=12At02415Bt05/2x(t)=v(t)dt+C2=(At23Bt3/2)dt+C2=12At2415Bt5/2+C2x(0)=0=C2sox(t0)=12At02415Bt05/2


a. a(t)=3.2m/s2t5.0sa(t)=1.5m/s25.0st11.0sa(t)=0m/s2t>11.0sa(t)=3.2m/s2t5.0sa(t)=1.5m/s25.0st11.0sa(t)=0m/s2t>11.0s;
b. x(t)=v(t)dt+C2=3.2tdt+C2=1.6t2+C2t5.0sx(0)=0C2=0therefore,x(2.0s)=6.4mx(t)=v(t)dt+C2=[16.01.5(t5.0)]dt+C2=16t1.5(t225.0t)+C25.0t11.0sx(5s)=1.6(5.0)2=40m=16(5.0s)1.5(5225.0(5.0))+C240=98.75+C2C2=−58.75x(7.0s)=16(7.0)1.5(7225.0(7))58.75=69mx(t)=7.0dt+C2=7t+C2t11.0sx(11.0s)=16(11)1.5(11225.0(11))58.75=109=7(11.0s)+C2C2=32mx(t)=7t+32mx11.0sx(12.0s)=7(12)+32=116mx(t)=v(t)dt+C2=3.2tdt+C2=1.6t2+C2t5.0sx(0)=0C2=0therefore,x(2.0s)=6.4mx(t)=v(t)dt+C2=[16.01.5(t5.0)]dt+C2=16t1.5(t225.0t)+C25.0t11.0sx(5s)=1.6(5.0)2=40m=16(5.0s)1.5(5225.0(5.0))+C240=98.75+C2C2=−58.75x(7.0s)=16(7.0)1.5(7225.0(7))58.75=69mx(t)=7.0dt+C2=7t+C2t11.0sx(11.0s)=16(11)1.5(11225.0(11))58.75=109=7(11.0s)+C2C2=32mx(t)=7t+32mx11.0sx(12.0s)=7(12)+32=116m

Additional Problems


Take west to be the positive direction.
1st plane: ν=600km/hν=600km/h
2nd plane ν=667.0km/hν=667.0km/h


a=vv0tt0a=vv0tt0, t=0,a=3.4cm/sv04s=1.2cm/s2v0=8.2cm/st=0,a=3.4cm/sv04s=1.2cm/s2v0=8.2cm/s v=v0+at=8.2+1.2tv=v0+at=8.2+1.2t; v=−7.0cm/sv=−1.0cm/sv=−7.0cm/sv=−1.0cm/s


a = −3 m/s 2 a = −3 m/s 2


b. t=7.8×108st=7.8×108s


1km=v0(80.0s)+12a(80.0)21km=v0(80.0s)+12a(80.0)2; 2km=v0(200.0)+12a(200.0)22km=v0(200.0)+12a(200.0)2 solve simultaneously to get a=0.12400.0km/s2a=0.12400.0km/s2 and v0=0.014167km/sv0=0.014167km/s, which is 51.0km/h51.0km/h. Velocity at the end of the trip is v=21.0km/hv=21.0km/h.


a = −0.9 m/s 2 a = −0.9 m/s 2


Equation for the speeding car: This car has a constant velocity, which is the average velocity, and is not accelerating, so use the equation for displacement with x0=0x0=0:x=x0+vt=vtx=x0+vt=vt; Equation for the police car: This car is accelerating, so use the equation for displacement with x0=0x0=0 and v0=0v0=0, since the police car starts from rest: x=x0+v0t+12at2=12at2x=x0+v0t+12at2=12at2; Now we have an equation of motion for each car with a common parameter, which can be eliminated to find the solution. In this case, we solve for tt. Step 1, eliminating xx: x=vt=12at2x=vt=12at2; Step 2, solving for tt: t=2vat=2va. The speeding car has a constant velocity of 40 m/s, which is its average velocity. The acceleration of the police car is 4 m/s2. Evaluating t, the time for the police car to reach the speeding car, we have t=2va=2(40)4=20st=2va=2(40)4=20s.


At this acceleration she comes to a full stop in t=v0a=80.5=16st=v0a=80.5=16s, but the distance covered is x=8m/s(16s)12(0.5)(16s)2=64mx=8m/s(16s)12(0.5)(16s)2=64m, which is less than the distance she is away from the finish line, so she never finishes the race.




v0=7.9m/sv0=7.9m/s velocity at the bottom of the window.


a. v=5.42m/sv=5.42m/s;
b. v=4.64m/sv=4.64m/s;
c. a=2874.28m/s2a=2874.28m/s2;
d. (xx0)=5.11×103m(xx0)=5.11×103m


Consider the players fall from rest at the height 1.0 m and 0.3 m.
0.9 s
0.5 s


a. t=6.37st=6.37s taking the positive root;
b. v=59.5m/sv=59.5m/s


a. y=4.9my=4.9m;
b. v=38.3m/sv=38.3m/s;
c. −33.3m−33.3m


h=12gt2h=12gt2, h = total height and time to drop to ground
23h=12g(t1)223h=12g(t1)2 in t – 1 seconds it drops 2/3h
23(12gt2)=12g(t1)223(12gt2)=12g(t1)2 or t23=12(t1)2t23=12(t1)2
0=t26t+30=t26t+3 t=6±624·32=3±242t=6±624·32=3±242
t = 5.45 s and h = 145.5 m. Other root is less than 1 s. Check for t = 4.45 s h=12gt2=97.0h=12gt2=97.0 m =23(145.5)=23(145.5)

Challenge Problems


a. v(t)=10t12t2m/s,a(t)=1024tm/s2v(t)=10t12t2m/s,a(t)=1024tm/s2;
b. v(2s)=−28m/s,a(2s)=−38m/s2v(2s)=−28m/s,a(2s)=−38m/s2; c. The slope of the position function is zero or the velocity is zero. There are two possible solutions: t = 0, which gives x = 0, or t = 10.0/12.0 = 0.83 s, which gives x = 1.16 m. The second answer is the correct choice; d. 0.83 s (e) 1.16 m


96km/h=26.67m/s,a=26.67m/s4.0s=6.67m/s296km/h=26.67m/s,a=26.67m/s4.0s=6.67m/s2, 295.38 km/h = 82.05 m/s, t=12.3st=12.3s time to accelerate to maximum speed
x=504.55mx=504.55m distance covered during acceleration
7495.44m7495.44m at a constant speed
7495.44m82.05m/s=91.35s7495.44m82.05m/s=91.35s so total time is 91.35s+12.3s=103.65s91.35s+12.3s=103.65s.

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