Question

quiz reviw l1 physics b
① what is vi from this graph?
② what is vf from this graph?
③ over the 5 seconds, what is the change in velocity?
④ what is the slope of the “change in velocity” line?
⑤ if something travels at -40m/s for 5 s, how far will it go & in what direction?
⑥ if something speeds up in a negative direction, is its acceleration + or -?
⑦ how far did an object thrown down at -40 m/s fall in 5 sec?

Explanation:

Step1: Identify initial velocity from graph

Initial velocity ($V_i$) is velocity at $t = 0$. From the graph, assume $V_i=20$ m/s (since no actual values on axes are clear - this is a placeholder for a general approach).

Step2: Identify final velocity from graph

Final velocity ($V_f$) at $t = 5$ s. Assume from graph $V_f=- 20$ m/s.

Step3: Calculate change in velocity

Change in velocity $\Delta V=V_f - V_i$. So, $\Delta V=-20 - 20=-40$ m/s.

Step4: Calculate slope of velocity - time graph

Slope of velocity - time graph gives acceleration. Using two - point formula $a=\frac{V_f - V_i}{t_f - t_i}=\frac{-20 - 20}{5-0}=-8$ m/s².

Step5: Calculate distance for constant velocity motion

For constant velocity $v=-40$ m/s and $t = 5$ s, distance $d=v\times t=(-40)\times5=-200$ m. The negative sign indicates motion in the negative direction.

Step6: Determine acceleration sign for negative - direction speeding up

If an object speeds up in the negative direction, its acceleration is negative because acceleration and velocity are in the same (negative) direction.

Step7: Calculate distance for falling object

Using the kinematic equation $d=v_i t+\frac{1}{2}gt^{2}$ with $v_i=-40$ m/s, $t = 5$ s and $g=- 10$ m/s².
$d=(-40)\times5+\frac{1}{2}\times(-10)\times5^{2}=-200 - 125=-325$ m.

Answer:

1. $V_i = 20$ m/s (assumed value based on general graph - reading concept)
2. $V_f=-20$ m/s (assumed value based on general graph - reading concept)
3. $\Delta V=-40$ m/s
4. Slope (acceleration) $=-8$ m/s²
5. Distance $=-200$ m, in negative direction
6. Acceleration is negative
7. Distance $=-325$ m