1. A particle moves along the x axis. Its position as a function of time is given by x=6.0t+8.5t2, where t is in conds and x is in meters. What is the acceleration as a function of time?
Solution:
We find the velocity and acceleration by differentiating x = (6.0 m/s)t + (8.5 m/s2)t2:
v = dx/dt = (6.0 m/s) + (17 m/s2)t;
a = dv/dt = 17 m/s2.
2. What is the centripetal acceleration of a child 3.6m from the center of a merry-go-round?The child’s speed is 0.85m/s.
Solution: The centripetal acceleration is
aR = v2/r = (0.85 m/s)2/(3.6 m) = 0.20 m/s2 ,direction: toward the center.
3. Huck Finn walks at a speed of 1.0m/s across his raft(that is,he walks perpendicular to the raft’s motion relative to the shore).The raft is traveling down the Mississipppi River at a speed of 2.5m/s relative to the river bank.What is the velocity(speed and direction)of Hack relative to the river bank?
Solution:
If vHR is the velocity of Huck with respect to the raft,
vHB the velocity of Huck with respect to the bank, and
vRB the velocity of the raft with respect to the bank, then
vHB = vHR + vRB , as shown in the diagram.
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From the diagram we get
vHB = (vHR2 + vRB2)1/2 = [(1.0 m/s)2 + (2.5 m/s)2]1/2 = 2.7 m/s.
We find the angle from
tan θ = vHR/vRB = (1.0 m/s)/(2.5 m/s) = 0.40, which gives θ = 22° from the river bank.
4. A 75-m-long train accelerates uniformly from rest.If the front of the train pass a railway worker 140m down the track at a speed of 25m/s,what will be the speed of the last car as it pass the worker?
Solution: We u a coordinate system with the origin at the
initial position of the front of the train. We can
find the acceleration of the train from the motion
up to the point where the front of the train pass
the worker:
v12 = v02 + 2a(D – 0);
(25 m/s)2 = 0 + 2a(140 m – 0), 小学生搞笑作文
which gives a = 2.23 m/s2.
Now we consider the motion of the last car, which starts at – L, to the point where it pass the worker:
v22 = v02 + 2a[D – (– L)]
= 0 + 2(2.23 m/s2)(140 m + 75 m), which gives v2 = 31 m/s.
5．what is the maximum speed with which a 1200-kg car can round a turn of radius 80.0
m on a flat road if the coefficient of friction between tires and road is 0.55?Is this result independent of the mass of the car?
Solution: If the car does not skid, the friction is static, with Ffr ≤ μsFN.
This friction force provides the centripetal acceleration. We take a
coordinate system with the x-axis in the direction of the
centripetal acceleration.
We write ∑F = ma from the force diagram for the auto:
x-component: Ffr = maR = mv2/R;
y-component: FN – mg = 0.
The speed is maximum when F小金蛋fr = Ffr,max = μ水利设计sFN.
When we combine the equations, the mass cancels, and we get
μsg = vmax2/R;
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(0.55)(9.80 m/s2) = vmax2/(80.0 m), which gives vmax = 21 m/s.
The mass canceled, so the result is independent of the mass.
6. A thin rod of length l carries a total charge Q distributed uniformly along its length.Determine the electric field along the axis of the rod starting at one end-that is,find E(x) for .
Solution: We choo a differential element of the
rod dx' a distance x' from the origin of
the coordinate system, as shown in the
diagram. the limits for x' are
0 to l. The charge of the element is dq = (少年先锋队队歌歌词Q/l) dx'. We find the
electric field by integrating along the rod:
7.林子祥的老婆 The resistance of a packing material to a sharp object penetrating it is a force proportional to the fourth power of the penetration depth,x:F=kx4i.Calculate the work done to force the sharp object a distanced.
Solution: The resisting force oppos the penetration. If we assume no acceleration, the applied force must be equal to this in the direction of the penetration. For a variable force, we find the work by integration:
8. A ball of mass 0.540kg moving east (+x direction) with a speed of 3.9m/s collides head-on with a 0.320-kg ball at rest.If the collision is perfectly elastic,what be the speed and direction of each ball after the collision?
Solution:
For the elastic collision of the two balls, we u momentum conrvation for this one-dimensional motion:
m1v1 + m2v2 = m1v1' + m2v2';
(0.540 kg)(3.90 m/s) + (0.320 kg)(0) = (0.540 kg)v1' + (0.320 kg)v2'.
广州市儿童公园Becau the collision is elastic, the relative speed does not change:
v1 – v2 = – (v1' – v2'), or 3.90 m/s – 0 = v2' – v1'.
Combining the two equations, we get
