HW #1 Solutions

(1-29)

(a) 1 mi = 1609 m = 1.609 km and 1 hr = 1 hr so 1 (mi/hr) = 1.609 (km/hr)

(b) 55 mi/hr *1.609 = 88.5 km/hr

(c) 65 mi/hr*1.609 = 104.6 km/hr ; D v=104.6-88.5 = 16.1 km/hr

(1-31) V = A t ® t = (V/A) = 3.78 ´ 10-3 m3 / 25 m2 =1.51 ´ 10-4 m

(2-6) x = 10 t2 so

t (s) à 2 2.1 3

x(m)à 40 44.1 90

  1. v = D x/D t = 50 m / 1 s = 50 m/s
  2. v = D x/D t = 4.1 m / 0.1 s = 41 m/s

(2-14) The answers are given by the slope of the graph

(a) v=0 (b) v<0 (c) v>0 (d) v=0

    1. (a) and (b) are graphs. Ask me in class or during office hours.
  1. The acceleration is the slope of the line at this point so a = -4 m/s2
  2. The position is given by the area under the v vs. t graph. It is easiest to break this up into part. From t=0 to t=3 "area" = 12; from t=3 to t=5 "area" = 16; from t=5 to t=7 "area" = 8; from t=7 to t=9 "area" = -8. At t=6 "area"=12+16+(8-2) = 34 so x=34 m2
  3. At t=9 "area"= 12+16+8-8 = 28 so x=28 m2

(2-31) V0 = 100 m/s a = -5.0 m / s2

V= V0 + a t à 0 = 100 -5.0 t à t= 20 s

V2= (V0)2 + 2 a x à 0 = (100)2 - 2*5.0 *x à x = 1000 m

No the runway is too short

(2-40) (a) t=(V- V0)/a = (5.4 ´ 105 m /s -3.0 ´ 105 m /s)/ 8.0 ´ 1014 m /s2 =3.0 ´ 10-10 s

(b) x= V0 t + a t2 /2=(3.0´ 105 m /s)( 3.0 ´ 10-10 s) -0.5*(8.0 ´ 1014 m /s2)( 3.0 ´ 10-10 s)

x = 1.26 ´ 10-4 m

(2-60) a = dv / dt = - 3 v2 and vo = 1.5 m/s. Solve this diff. Eqn.

v t

ò vo (1/v2 ) dv = - 3 ò o dt à -(1/v) + (1/ vo) = -3 t we want v = vo/2

so t = (-1/3) *[-(1/v) + (1/ vo)] = -(1/3) *[-1/ vo] = 0.22 seconds

(2-64) (a) x-xo =40 m and vo =0 and v = 0.01(3.0 ´ 108 m /s)= 3.0 ´ 106 m /s so using Eq. 2.12 gives

a=(v2 - vo2 )/[2(x - xo )] = 1.12 ´ 1011 m /s2

(b) The time for the first part of the flight is given by using Eq, 2.11

t1= Ö (2*( x-xo)/a) = 2.67 ´ 10-5 s. For the second part of the flight (where a=0) we find

t2 = (distance)/velocity = 60 m / 3.0 ´ 106 m /s =2.0 ´ 10-5 s

The total time is just the sum of these so ttot = t1 + t2 = 4.67´ 10-5 s

(2-66) (a) Using Eq. (2.12) gives a=(v2 - vo2 )/[2(x - xo )] = (125)2 / 2*250 = 2.41 m /s2

(b) Using Eq. 2.8 gives t=(v- vo)/a = 34.7/2.41 = 14.4 s

(c ) Using Eq. 2.8 again v = (2.41 m /s2)(25 s) = 60.25 m /s

(2-74) One the way down the rock undergoes constant acceleration so the relationship between time and distance is d=(1/2)*g*(t1)2 = (1/2)*(9.8)*(t1)2 . On the way back up the sound wave travels at constant velocity so d=336*t2 . t1 and t2 are the time it take the rock to fall down the well and the time it takes the sound to travel back up the well respectively. The total time t1 + t2 =2.4 Now d = 336* t2 =(1/2)*g*(t1)2=(1/2)*g*(2.40-t2)2 (where in this last step I used t1 + t2 =2.4 to replace t1 ). This gives the following equation for t2 : 336* t2 =4.9*(2.40-t2)2. Expanding out the square and rearranging terms gives the following quadratic equation for t2 : 4.9 (t2 )2 -359.5* t2 +28.22=0 . Solving this and taking the positive root gives t2 = 0.0765 sec. So d=336* t2 =26.4 m.

(b) If the sound travel time is ignored then d= (1/2)*9.8*(2.4)2 =28.22 m, which is an error of about 6.88%

(2-80) (a) In walking a distance D x, in a timeD t, the length of rope l is only increased by D x(sinq ) where q is the angle between l and the vertical h. (This is a bit tricky to figure out. Just draw a right triangle near the boys hand). So the rate at which the pack is lifted is v =D x(sinq )/ D t = vo (x/l) (since D x/ D t = vo and sinq = x/l ). So v= vo[x/Ö (x2 + h2 ) ].

(b) Now a = dv/dt = (vo/l)(dx/dt) +(vo x)(d (1/l)/ dt) = (vo/l)( vo) - (vo x/l2 )( d l / dt). Now d l / dt= vo so a = (vo2/l) [ 1 - (x2 / l2 ) ] = (vo2/l) [ h2 / l2 ] = vo2 [ h2 / l3 ]

So a = vo2 [ h2 / (x2 + h2 )3/2 ].

(C ) Using the results from (a) and (b) and taking the limit xà 0 we get v=0 ; a = vo2/h

(D) Using the results from (a) and (b) and taking the limit xà ¥ we get v=vo ; a = 0