Then time can be obtained by using 700=69.8 t.

Problems (Measurements, Vectors, Kinematics, Laws of Motion)
1. The position vector of an object is given by
in meters.
a)
b)
c)
d)
, where d, b and c are constants and t is given in secs and r is
What is the units of d, b and c?
What is the displacement of the object between t=0 and t=4 sec?
What is the velocity and acceleration of the object at t=2 sec?
If d=1, b=1/2 and c=-1, Find the angle between a and r.
2. A dog (Newton's dog) sees a flowerpot sail up and then back down a window (height of window is 1.1m). If the total time the
pot is in sight for 0.54 s, find the height above the top of the window to which the pot rises. (Halliday and Resnick)
Solution hints:
2
The pot passes the window t=0.54/2=0.27 sec. Acceleration g=9.8 m/s .
2
Velocity of pot at the top of window can be calculated from the equation: h=v 0t+1/2gt .(v0=2.75 m/s).
2
2
h=1.1 m. Then v0=5.4 m/s. Then using (vf) =(v0) -2gy; final velocity is zero at the top. Then y=0.39 m.
3. A student is running at her top speed of 5.0 m/s to catch a bus, which is stopped at the bus stop. When the student is still 40.0 m
2
from the bus, it starts to pull away, moving with a constant acceleration of 0.170 m/s .
(a) For how much time and what distance does the student have to run at before she overtakes the bus?
(b) When she reaches the bus, how fast is the bus traveling?
(c) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student
and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How
fast is the bus traveling at this point?
(d) If the student’s top speed is 3.5 m/s will she catch the bus?
(e) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she
have to run in that case? (Young&Freedman)
Hints:
At t=0, the distance between bus and student is 40 m. Position of student is:
xs=-40+5t.
Position of the buss is
2
xb=1/2at .
(a) When the student catch the bus xb=xs then t=9.55 sec and 49.27 sec. (two roots)
Answer is positive root (9.55 sec). the student run 47.8 m.
(b) v=at=1.6 m/s.
(c)At t=9.55 sec. student still faster than the bus. If she continue running she pass over the bus but buss is accelerating. Then they
meet at t=49.27 sec again. At this point speed of buss is 8.4 m/s.
Let us answer part (e) before part (d).
(e). Minimum speed should be 3.7 m/s. (try yourself to the other parts)
o
4. A certain airplane has a speed of 290.0 km/hr and is diving at an angle of 30.0 below
horizontal when the pilot releases a radar decoy. The horizontal distance between the
release point and the point where the decoy strikes the ground is 700 m. (a) How long is
decoy in the air? (b) How high was the release point? (see figure)
the
the
0
Hint: Initial velocity of decoy is 290 km/hr with an angle of 30 below the horizontal.
Horizontal component of the velocity is
Then
can be obtained by using 700=69.8 t.
time
5. A batter hits a pitched ball when the center of the ball is 1.22 m above the ground. The ball leaves the bat at an angle of 45o
with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107 m. (a) Does the
ball clear a 7.32 m–high fence that is 97.5 m horizontally from the launch point? (b) Either way, find the distance between the
top of the fence and the center of the ball when the ball reaches the fence.
6. A 0.5 kg puck rest on a frictionless surface at the origin of the coordinate system. Two string are attached. One is pulled with a
force of 2.5 N along the y-axis and the other with a force of 6.0 N along the x-axis.
a)
b)
c)
d)
e)
Where is the puck 10 secs later?
What is the pucks velocity at this time?
What is the work done by the force along x-axis on the puck during this time?
What is the work done by the force along y-axis on the puck during this time?
What is the total work done by the forces?
7. An elevator and its load have a combined mass of 1600 kg. Find the tension in the
supporting cable when the elevator, originally moving downward at 20 m/s, is brought to
rest with constant acceleration in a distance of 50 m.
What will be the tension in the cable when the elevator moving upward with the same
constant acceleration and load?
8. Angle for Minimum Force. A box with weight is pulled at constant speed along a level
floor by a force F that is at an angle above the horizontal. The coefficient of kinetic friction between the floor and box is
(a) In terms of
,
and
.
, calculate F.
(b) From the general expression in part (a), calculate the value of
a minimum. (Hint: At a point where a function is minimum, what
for which the value of F, required to maintain constant speed, is
are the first and second derivatives of the function? Here F is a function of
evaluate this optimal .
) For the special case of
N and
9. A block slides down an inclined plane of slope angle A with constant velocity. It is then projected up the same plane with an
initial
speed
Vo.
a)
How
far
up
the
incline
will
it
move
before
coming
to
rest?
b) Will it slide down again?
10. A small object of mass m =0.2 kg is placed at the top of a large sphere of radius R =0.5 m resting on the ground. The object is
given a negligibly small velocity so that it starts to slide down the sphere. Assume the surface of the sphere is frictionless and the
sphere is fixed to the surface of a table. In this problem, you will try to find where the object hits the ground.
a) Briefly describe how you intend to model the motion. Here are some questions which should help guide your thinking. Is
there any special condition(s) that describe when the object leaves the sphere? Does the normal force do any work on the
object? Does the principle of Conservation of Energy replace Newton’s Second Law in the radial direction or are they
independent?
b) At what angle will the mass leave the sphere?
c) What is the velocity of the mass when it just leaves the sphere?
d) How far from the bottom of the sphere will the mass hit the ground?
11. Tension in an elavator cable. An elevator and its load have a combined mass of 800 kg (Fig.). The elevator is initially moving
downward at it slows to a stop with constant acceleration in a distance of 25.0 m. What is the tension T in the supporting cable
while the elevator is beingbrought to rest? (Ans: T=9440 N)