[BS1102] General Physics
Session 2 - Kinematics: 1D Motion
Homework Assignment
A.
Name
Student ID
Class
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Acceleration and Freely Falling Objects
1) A car moving in a straight line starts atx = 0 at t = 0. It passes the point x = 25.0 m with a speed of 11.0
m/sat t = 2.0 s. It passes the point x = 385 m with a speed of 45.0 m/sat t = 25.0 s. Find (a) the average
velocity, and (b) the average acceleration, between t = 2.00 sand t = 25.0 s.
a. x = 25 s = 11m/s t = 2 s
x2 = 385 m s2 = 45 m/s t2= 25 s
a = 385-25/25-2 = 15.652 m/s
b. 45-11/25-2 = 1.478m/s
2) A stone is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an
initial height of 50.0 m above the ground. The stone just misses the edge of the roof on its way down.
Determine:
(a) The time needed for the stone to reach its maximum height.
(b) The maximum height.
(c) The time needed for the stone to return to the height from which it was thrown and the velocity of
the stone at that instant.
(d) The time needed for the stone to reach the ground.
(e) The velocity and position of the stone at t = 5.00 s.
c.t = 2 x 2.04
v = 20 m/s h0 = 50 m
= 4.08 s
a.t = 20/9.8 = 2.04 s
v=20m/s
b. vo2/2g + h
d. y=yo+voxb400/19.6 + 50
1/2gt2
=20.4 + 50
-50= 0 +
=70.4 m
20x-4.9t2
-50
t = quadratic fornula
t= 5.83 s
B.
Graphical Analysis of Motion
3) The position-time graphs for two i3L students (A and B), returning from campus, to their homes at P
and Q respectively,are shown in the following figure. Determine if each statement is corrector not and
briefly explain your reasoning. Use the kinematic equations if necessary.
(a) A and B walks to the opposite direction from
each other.
(b) A walks at a constant speed but B walksatan
increased speed
(c) A lives closer to campus than B.
(d) A starts from the campus earlier than B.
(e) A walks faster than B.
(f) A reaches home earlier than B reaches his
home.
a. No Because there is no negative position
b. No because both go at constant speed
c. Yes because because B left after A
d. Yes because A left a t=0 but B is after A
e. No because A reaches the same time as B eventhough he left first
f. No they arrive at the same time because both in the graph stop at the same t
4) Based on the velocity vs graph below, calculate:
(a)
(b)
(c)
(d)
(e)
The acceleration from t = 0 sto t = 15 s.
The average acceleration from t = 0 sto t = 30 s.
The displacement from t = 0 sto t = 15 s.
The displacement from t = 0 sto t = 30 s.
(Challenge) The average velocity from t = 0 s to
t = 15 s.
(f) (Challenge) The average velocity from t = 0 s to
t = 30 s.
END OF HOMEWORK