Question

name: motion and forces study guide
equations: familiarize yourself with the following equations
$v = \frac{d}{t}$
$slope=\frac{rise}{run}=\frac{y_2 - y_1}{x_2 - x_1}$
$a=\frac{f_{net}}{m}$
$f_g = mg$
$a=\frac{v_f - v_i}{t}$
$g=-9.8\frac{m}{s^2}$
$p = mv$
motion:

1. define: distance, displacement, speed, velocity, and acceleration.
2. of the terms listed above, which are vector quantities? which are scalar quantities? whats the difference between the two?
3. what is the slope of a position vs time graph (see vocab in #1)?
4. what is the slope of a velocity vs time graph (see vocab in #1)?
5. what are the three things an object could be doing if it is accelerating?
6. use the position vs time graph below to answer the following questions.

a. what is the objects position at $t = 25s$?
b. what is the objects velocity at $t = 25s$?
c. what is the objects velocity from $t = 0s$ to $t = 10s$?
d. what is the objects velocity from $t = 10s$ to $t = 15s$?
e. what is the objects velocity from $t = 15s$ to $t = 40s$?
f. when is the object at rest?
g. when is the object moving to the left (negative velocity)?

Explanation:

Step1: Define terms

Distance is the total path length traveled. Displacement is the straight - line distance from the starting to the ending point with direction. Speed is the rate of change of distance ($v=\frac{d}{t}$), a scalar. Velocity is the rate of change of displacement, a vector. Acceleration is the rate of change of velocity ($a = \frac{v_f - v_i}{t}$).

Step2: Identify vector and scalar

Vector quantities (have magnitude and direction): displacement, velocity, acceleration. Scalar quantities (have only magnitude): distance, speed. The difference is that vectors need direction information while scalars don't.

Step3: Slope of position - time graph

The slope of a position - time graph is velocity. Using the slope formula $slope=\frac{y_2 - y_1}{x_2 - x_1}$, for a position ($y$) vs time ($x$) graph, it gives velocity as $v=\frac{\Delta x}{\Delta t}$.

Step4: Slope of velocity - time graph

The slope of a velocity - time graph is acceleration. Using the slope formula on a velocity ($y$) vs time ($x$) graph gives $a=\frac{\Delta v}{\Delta t}$.

Step5: Things an object can do while accelerating

1. Changing speed (speeding up or slowing down). 2. Changing direction while moving at constant speed. 3. Changing both speed and direction.

Step6a: Object's position at $t = 25s$

From the graph, at $t = 25s$, the position is $60m$.

Step6b: Object's velocity at $t = 25s$

From $t=20s$ to $t = 30s$, the object is at rest (constant position), so velocity at $t = 25s$ is $0m/s$.

Step6c: Object's velocity from $t = 0s$ to $t = 10s$

$v=\frac{\Delta x}{\Delta t}=\frac{60 - 0}{10-0}=6m/s$.

Step6d: Object's velocity from $t = 10s$ to $t = 15s$

The object is at rest (constant position), so velocity is $0m/s$.

Step6e: Object's velocity from $t = 15s$ to $t = 40s$

From $t = 15s$ to $t=30s$, velocity is $0m/s$. From $t = 30s$ to $t = 40s$, $v=\frac{\Delta x}{\Delta t}=\frac{0 - 60}{40 - 30}=-6m/s$.

Step6f: Object at rest

The object is at rest from $t = 10s$ to $t = 15s$ and from $t = 20s$ to $t = 30s$.

Step6g: Object moving to the left

The object is moving to the left (negative velocity) from $t = 30s$ to $t = 40s$.

Answer:

1. Distance: total path length; Displacement: straight - line distance with direction; Speed: rate of change of distance (scalar); Velocity: rate of change of displacement (vector); Acceleration: rate of change of velocity.
2. Vector: displacement, velocity, acceleration; Scalar: distance, speed. Difference: vectors have direction, scalars don't.
3. Velocity.
4. Acceleration.
5. Changing speed, changing direction, changing both speed and direction.

6a. $60m$
6b. $0m/s$
6c. $6m/s$
6d. $0m/s$
6e. From $t = 15s$ to $t = 30s$: $0m/s$, from $t = 30s$ to $t = 40s$: $-6m/s$
6f. From $t = 10s$ to $t = 15s$ and from $t = 20s$ to $t = 30s$
6g. From $t = 30s$ to $t = 40s$