普通物理学(第五版)力学运动答案
1-1解:(2)质点在1秒到3秒时间内的平均速度为:v=(3)由作图法可得到质点在t =0时的位置为:x=2.71m
3.75−3.003.0−1.0
=0.285(m/s)
1-2解:x=4t-2t3 (1)∆x =x −0=4t −2t3=4×2−2×23=−8 v ==
s∆t
dxdt
m∆x−82
=−
4ms
v =
=4−6t2=4−6×22=−
44ms
20ms
(2)∆x =x3−x2= 4×3−2×33 − 4×1−2×1 =
2ms
−v =
∆x∆t
=
−443−1
=−
22ms
(3)v1=4−6t2=−v3=4−6t2=−50m/s α =
v3−v1t3−t1
=
−24m/s (4)α=
dvdt
=−12t=−36m/s
12
1-3解:由v~t图的总面积可得到路程为:S=1/2(30+10)×5+ 20×10 =200m 总位移为:∆x =2 30+10 ×5−2 20×10 =0所以平均速度也为零 1-4解:v0=b
ms
1
1
, t1=2b s, v0=0(1)求B 在时刻t 的加速度。在v′~t坐标系中质点2的运
动方程为:v′2+t2=(v0+c) 2因为v′=v+c在v~t坐标系中质点2的运动方程为:
(v+c) 2+t2=(v0+c) 2……(1)当t=2b,v=0,v0=b代入式(1)得:c=3/2b,∵v0=b代入(1)化简后得:v2+3bv+t2=4b2……(2)解得:v =α=
dvdt
−3b± 2
对t 求导后得:
=2b
当t=2b时B 静止A 追上B ,A 的位移等于B 的位移B 的位移:
3b2
∆xb= vdt= 0(− 0 dt
2b
+1/2 dt=−3b2+ 01/2 dt其中:=
252
2b
b2 2+1/2sin ϑcos ϑ arcsin 5=8.79b2 ∆xB=3b2+0.5×
2b
ϑ4
8.79b2=1.40b2设A的速度为:vA=kt ∆xA= vdt= 0ktdt=2kb2相遇时A 与B 的位移相等:∆xA=∆xb 1.40b2=2kb2 k=0.7 vA=kt=0.7t αAvA=vB时有:0.7t =−1-5解:
ℎ
l
x+b
ℎ
3b2
dvAdt
=0.7m/s (3)当
+ t =1.07b
2
dbdtlℎ−l
1
= ∴ℎb=(x+b) 上式两边微分得到:h
dbdt
=l
d(x+b) dt
=l
dxdt
+l
dbdt
而
dxdt
=v0影
子长度增长速率为:
d (x+b)
dt
=
lℎ−l
v0∵ℎb=(x+b)
dbdt
=
v0所以人影头顶移动速度为:
=
ℎdbldt
=ℎ−lv0 =−v0
y=y0−v0t x2+y2=l2将此式微分得:2ydy+2xdx=0
dxdt
ℎ
1-6解:t=0
y
y=y0=4
dydt
=−xdt=
ydy
−x −v0 =
00y0=4v0=2 t=1代入,得B 端的速度:0.87m/s x = dx=
+c=+c= +c t=0 x= +
c=x0=3∴c=0 x= 1-7解: =x −ℎ =
r
i
jv
d
r
d
dt
dt
=
dx
dti
=r=dt =
r
r
d
v
=
=−v0
=
v
d
r
dt
=
dx
=−dt
i
v0x
=
ia
dt
=
d2x
=−dti
v02ℎ2x
i
1-8解:dy =−bk e−ktdt y= dy= −bke−ktdt+c=be−kt+c当t=0 y t=0=b+c=
22dxx=aektktdx2ktdy2−kt
b∴c=0 轨迹方程: =−kt xy=ab dt=akedt=akedt=bkeay=be
ak2ekt +bk2e−kt
i
j
1-9解: =
v
d
dt
=8t + =
j
ka
d
dt
=8 x=1 y=4t2 z=t轨迹方程:y=4z2 x=1轨迹为在
j
x=1平面的一条抛物线。
1-10解:(1) =(3t+5) +(1/2t+3t−4) (2)y =
2
r
i
j
73
j
i
j
2
1x−52
3
+3
x−53
−4 (3) =3 +(t+
v
i
3) =3 +7 tan α= α=66.80 v= =7.61m/s 1-11解:由y2=2x两边微分得:ydy =2dx
1+y y′′
3′231 1+
y−dydx
=
2
1y
=tan α
d2ydx2
=−
1dyy2
dx
=
1y3
R=
2 2
3
= =(1+y)
23/2
mg(2-y)=1/2mvmgcos α−N=
mv2R
R= 1+y
cos α=
=
N=0 得:y3+3y−4=0 (y-1)(y2+y+4)=0其中(y2+y+
12
4)=0有两个虚根,不符题意。∴y1=1, x= 1-12
dvdty
解
dydvdsdt
:
dyds
=−g sin θ sin θ==
dvdsdsdt
=−g sin θ=−gv
dvds
=−g
dyds
vdv=-gdy
v0vdv=
v
y−gdy v2−v02=2g(y−y0)
1-13
解:θ=θ0+wt=wt x=ℎtan θ=ℎtan wt v=
dxdt
=ℎwsec2wt α=
d2xdt=
2ℎw2sec2wttan wt
x=vtLcos θ=vt2v2sin θ
1-14解: y=1/2gt2 L==270m
gcos2θLsin θ=1/2gt2
2x=v0cos θ轨迹方程为:y =x tan θ−gx1-15解: =xtan θ−
2v0cosθy=v0sin θt−1/2gt2
gx22v02
1+tanα …… 1 由=0 x−2v22tan θ=0得:tan θ=
dtan θ
2
dy
gx2
v02gx
(1)可得:
y =x gx−2v2−2v
v02
gx2
gx2v04
02
g2x2
=
v022g
−2v2=12.3m
gx2
1-16解:由轨迹方程:y =x tan α−2v
gx2
2cos2θ
=xtan α−2v2 1+tan2α
gx2
2v02ygx2
=
2v02gx
tan α−
1+tan2α tan2α− 1−v2(y+2v)
2v02gx
tan α+ 1+
2v02ygx =0tan α=
2v022v2vy
± (0) 2−4(1+0=
v02gx
1±
2ggx2x=v0t22v02ℎ2R2w2ℎ
=R 1+2wℎ1-17解:v0=Rw x== r=gggy=1/2gt2
x=0x′=−5td2xd2y1-18解:s 系: S ’系: ′s 系:ax= dt=0 ay= dt=
y=v0t−1/2gt2y=v0t−1/2gt2g =−g S ’系:a′x=
a
j
d2xdt=0 a′y=
d2ydt=g =−g
a
j
1
1-19解:乙船的人看甲:v′1= =11.2m/s a1=arctan =26.60甲船的人看乙:
2v′2= =11.2m/sa1=arctan 2 =63.40 1-20解:(1)l =v ′sin αt t=
lv′sinα
10001.5×sin15==2546s L=( v−v′cos150)t =(2-1.5×
lv′
cos150) ×2564=1.41km (2)欲使时间最短α′=900l =v′t t=1.33km (3) l=v′sinθt L2=(v−v′cosθ )tt=
dL2dθ2
lv′sinα
=
10001.5
=667sL1=vt=
lv′sinα
L2=(v−v′cosθ )
令:
=
ddθ2
(v−v′cosθ )
′2
2
lv′sinα
2
=0
′
得: l=
(v−v′cosθ )lv′cosθ
v′2sin2θ
∴v′2sin2θ=vv′cosθ−
v′cosθ vcosθ+sinθ = vvcosθcosθ=
lv′sinα
v′2vv=
22.53
=0.75 θ=41.40L2=
lv′sinα
v−v′cosθ = 2−1.5×cos41.40
10001.5×sin41.4=0.89kmt==
10001.5×sin41.4=
1010s 1-21解:(1)假定空气是静止的(即vr= 0 ),试证来回飞行时间为,t0 =2 l/v′。当:vr= 0 t=t0 =2 l/v′(2)往程:v1=v′+vr t往=
lv′+vrlv2
lv+vr
返程:v2=v′−vrt返=
lv1
lv−vr
t= t往+t返=
+
lv′−vr=
2
1−()
t0
(3)往程:v1= v′2−vr2 t往=
v′−vr2
=
v′−vr2
v2= v′2−vr2t返=
=
v′−vr2
t= t往+t返=
=
t0
1−(2
vr
1-1
1-2(1) 1-2(2)1-51-6
1-61-71-111-12
1-131-141-161-171-19(1)
1-19(2)1-201-21