西南交大 大学物理 英文 试题 答案No.A1-2.11348895

average velocity is 0.0068iˆ + 0.0025 ˆj(m/s) , and the magnitude of its average velocity is
7.24 ×10−3 m/s .
Solution:
(a) The coordinate system is shown in figure. The initial position vector is rv = −0.15iˆ , and the
4. The flywheel of an engine I rotating at 25.2 rad/s. When the engine is turned off, the flywheel decelerates at a constant rate and comes to rest after 19.7s. The angular acceleration (in rad/s2) of the flywheel is 12.8 rad/s2 , the angle (in rad) through which the flywheel rotates in coming to rest is 248.1 rad , and the number of revolutions made by the flywheel in coming to rest is 39.5 . Solution:
answer.
3. An object is launched into the air with an initial velocity given by vr0 = (4.9iˆ + 9.8 ˆj)m/s .
Ignore air resistance. At the highest point the magnitude of the velocity is
3. The angle turned through by the flywheel of a generator during a time interval t is given by
θ = at + bt 3 − ct 4 , where a, b, and c are constant. Its angular velocity is a + 3bt 2 − 4ct 3 . And
[ ] [ ] ⇒ rv(t) ⋅ [ωv(t) × rr(t)] = r cosθ (t)iˆ + r sinθ (t) ˆj ⋅ r cosθ (t)ωz (t) ˆj − r sinθ (t)ωz (t)iˆ
= r 2 sinθ (t) cosθ (t)ωz (t) − r 2 sinθ (t) cosθ (t)ωz (t) = 0 (2) According to ωr(t) × rr(t) = vv(t) and rr(t) ⋅[ωr(t) × rr(t)] = 0 , we know rr(t) ⋅ vv(t) = 0 . The result means that vr(t) is always perpendicular to rr(t) .
Branch Fig.1
SI units using the crook in the branch as the origin for a set
of horizontal and vertical coordinate axes.
(b) If the caterpillar traverses the distance during 1.0 min, its average speed is 0.0075m/s , its
E2ωr.x(Uptl)saei=nEωqwuzha(ytti)oktˆnhi4st.o2r8eesvfuaollrtuairrste(tat)hne:otehrrxe(prtr)ews=saiy[orntcoorrss(eθte)(⋅tt[)hω]ariˆt(+t )vr[×(rtrsr)i(nti)sθ]
University Physics AI
No. 2 Motion in Two and Three Dimensions
Class
Number
Name
I．Choose the Correct Answer
1. An
t =0
object moves in the object has
the xy plane with a velocity given
(1) According to the equation 4.49 ω(t) = ω0 + αt , the angular acceleration is
α = − ω0 = − 25.2 = −1.28 rad/s2 t 19.7
(2) According to the equation of motion
y 30cm
branch inclined at 30o to the horizontal as shown in Figure
15cm
30o
x
1. (a) The initial position vectors − 0.15iˆ and the final
0
position vectors 0.15 3iˆ + 0.15 ˆj of the caterpillar in
(t)] ˆj and Equation 4.26
=
0.
always perpendicular to
for
rr (t
ωrr motion: solution(2) .
Solution:
{ } (1)ωv(t) × rr(t) = ωz (t)kˆ × [r cosθ (t)]iˆ + [r sinθ (t)] ˆj = r cosθ (t)ωz (t) ˆj − r sinθ (t)ωz (t)iˆ
an by
avrcc=ele3riˆat+io0nˆjth. aWt hhaast
a positive x component. At time can be concluded about the y
component of the acceleration?
(D)
(A) The y component must be positive and constant.
(B)
(A) 0. (B) 4.92 m/s . (C) 9.82 m/s . (D) (4.9)2 + (9.8)2 m/s .
Solution: The y component of the velocity is zero at the highest point, and the x component of the velocity is
ar . Which of the following expression are also
constant?
d vr
(A)
.
dt
dvr
(B) .
dt
d(v2 )
(C)
.
dt
d(vr/ vr )
(D)
.
dt
(B)
Solution: According to the relation of the acceleration and the velocity av = dvv , we know the dt
of the car when t=0s is 6.0m/s2 , the magnitude of the centripetal acceleration of the car when t=5.00s is 3.375m/s2 , the magnitude of the angular acceleration is 0.01rad/s2 , the magnitude of the tangential acceleration is 1.5m/s2 .
its angular acceleration is 6bt − 12ct 2 .
Solution:
Its angular velocity is ω = dθ = d (at + bt 3 − ct 4 ) = a + 3bt 2 − 4ct 3 dt dt
And its angular acceleration is β = dω = d (a + 3bt 2 − 4ct 3 ) = 6bt − 12ct 2 dt dt
vvave
=
Δrv Δt
=
rvf − rvi Δt
=
0.26iˆ + 0.15 ˆj + 0.15iˆ 60
= 0.0068iˆ + 0.0025 ˆj m/s.
The magnitude of the average velocity is
vvave = 0.0068iˆ + 0.0025 ˆj = 0.00682 + 0.00252 = 7.24 ×10−3 (m/s)
(B) The y component must be negative and constant.
(C) The y component must be zero.
(D) Nothing at all can be concluded about the y component.
Solution: According to the definition of the acceleration components
5．A car is traveling around a banked, circular curve of radius 150 m on a test track. At the instant when t=0s, the car is moving north, and its speed is 30.0 m/s but decreasing uniformly, so that after 5.0 s its angular speed will be 3/4 that it was when t=0s. The angular speed of the car when t=0s is 0.2rad/s , the angular speed 5.0 s later is 0.15rad/s , the magnitude of the centripetal acceleration