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Chapter 2
Motion in One Dimension
Problem 1
A particle moves along the x axis according to the equation x = 2.01 + 2.97 t - t^2 , where x is in meters and t is in seconds.
(a) What is its velocity at t = 2.80 s?
(b) What is its acceleration at t = 2.80 s?
Kinematic Equations
The kinematic equations may be used to
solve any problem involving one-dimensional
motion with a constant acceleration
You may need to use two of the equations to
solve one problem
Many times there is more than one way to
solve a problem
Kinematic Equations, specific
For constant a ,
Can determine an object’s velocity at any
time t when we know its initial velocity and its
acceleration
Does not give any information about
displacement
v xf = vxi + a tx
Kinematic Equations, specific
For constant acceleration,
The average velocity can be expressed as
the arithmetic mean of the initial and final
velocities
2
xi xf x
v v v
=
Kinematic Equations, specific
For constant acceleration,
This gives you the position of the particle in
terms of time and velocities
Doesn’t give you the acceleration
x f = xi + vt = xi + vxf + vxi t
Kinematic Equations, specific
For constant acceleration,
Gives final position in terms of velocity and
acceleration
Doesn’t tell you about final velocity
(^12)
2
f i xi x
x = x + v t + a t
Kinematic Equations, specific
For constant a ,
Gives final velocity in terms of acceleration
and displacement
Does not give any information about the time
2 2 2 ( ) f i f i v = v + a x − x
Problem 3
The position versus time for a certain particle moving
along the x axis is shown in the figure below.
Find the average velocity in the following time intervals.
(a) 0 to 2 s
(b) 0 to 4 s
(c) 2 s to 4 s
(d) 4 s to 7 s
(e) 0 to 8 s
Problem 4
A position-time graph for a particle moving along the x axis is shown in the figure. The divisions along the horizontal axis represent 1.50 s and the divisions along the vertical axis represent 5.0 m. Determine the instantaneous velocity at t = 6.00 s (where the tangent line touches the curve) by measuring the slope of the tangent line shown in the graph.
Freely Falling Objects
A freely falling object is any object moving
freely under the influence of gravity alone.
It does not depend upon the initial motion of
the object
Dropped – released from rest Thrown downward Thrown upward
Acceleration of Freely Falling
Object
The acceleration of an object in free fall is directed downward, regardless of the initial motion
The magnitude of free fall acceleration is g = 9. m/s^2 g decreases with increasing altitude g varies with latitude 9.80 m/s^2 is the average at the Earth’s surface
Acceleration of Free Fall, cont.
We will neglect air resistance
Free fall motion is constantly accelerated
motion in one dimension
Let upward be positive
Use the kinematic equations with ay = g = -
9.80 m/s^2
Free Fall Example
A ball is thrown vertically upward from the
ground with an initial speed of 15.0 m/s. (a)
How long does it take the ball to reach its
maximum altitude? (b) What is its maximum
altitude? (c) Determine the velocity and
acceleration of the ball at t = 2.00 s.
Answer: (a) 1.53 s; (b) 11.5 m;
(c) v = –4.60 m/s, a = –9.80 m/s^2
General Problem Solving
Strategy
Conceptualize
Categorize
Analyze
Finalize
Problem Solving –
Conceptualize
Think about and understand the situation Make a quick drawing of the situation
Gather the numerical information Include algebraic meanings of phrases Focus on the expected result Think about units
Think about what a reasonable answer should be
Problem 5
A rock is dropped from rest into a well. The
sound of the splash is actually heard 2.4 s
after the rock is released from rest. How far
below the top of the well is the surface of the
water? The speed of sound in air (at the
ambient temperature) is 336 m/s.
