"Uniform" means "constant." If a is constant and not zero, then x = u.t does not apply!
Quantities of Interest
Time (t) - Clock reading. t and Dt are always positive.
Position (x or y or z) - An object's location with reference to the origin of a coordinate system.
Displacement
- Change in position. For one-dimensional motion along the x-axis,
displacement is: Dx
= x- xo.
The expression "distance moved" is more or less synonymous with
displacement.
You will need to look at the wording and the context of the problem.
Velocity
(u)
- The instantaneous velocity at any time t is the slope of the x
vs. t graph at that point in time.
Our text uses uo
to indicate initial velocity and u
to indicate final velocity.
Acceleration (a) - The instantaneous acceleration at any time t is the slope of the u vs. t graph at that point in time.
The magnitude of a quantity is its absolute value, always a positive number.
Test
yourself (1).
What is the algebraic sign of the velocity of an object under these
conditions?
(a) On the
positive x-axis and moving toward the origin.
(b) On the
negative x-axis and moving away from the origin.
(c) On the
negative x-axis and moving toward the origin.
(d) On the
positive x-axis and moving away from the origin.
Check
your
answers.
Positive
acceleration means
either u
is in the + x direction and is increasing in magnitude,
or u
is in the -x
direction and is decreasing in magnitude.
Negative
acceleration means
either u
is in the + x direction and is decreasing in magnitude,
or u
is in the -x
direction and is increasing in magnitude.
Test
yourself (2). For each of these cases, what is the sign of the
acceleration?
(a) uo
= -2
m/s, u
= -1
m/s.
(b) uo
= +3 m/s, u
= 0 m/s.
(c) uo
= -2
m/s, u
= -4
m/s.
(d) uo
= +2 m/s, u
= +6 m/s.
Check
your
answers.
Basic
Relationships
A Good Problem-Solving Strategy
Identify the known variables and the missing quantities. |
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1. A truck moving at 12.0 m/s brakes uniformly to a stop in 8.30 seconds. Calculate the distance that the truck travels while braking.
Given: uo = 12 m/s, u = 0, t = 8.3 s. Missing: a, x. Find: x.
Sketch a
graph
of u
vs. t.
2. A ball is thrown into the air and rises to a maximum height of 4.5 m. How much time does it take for the ball to return to a point 2 m below its starting point?
Given: ymax = 4.5 m, yfinal = -2 m, a = -9.8 m/m/s2. Missing: uo, ufinal, t.
Sketch a u
vs. t graph of the motion
Test yourself (1) Answers: (a) and (b) negative, (c) and (d) positive.
Test yourself (2) Answers: (a) positive, (b) and (c0 negative, (d) positive.