Question

1. the displacement of an object is plotted along the time. describe the velocity and the acceleration of the object. 2. the motion of a car along a straight line is shown in this position vs time graph. what is the velocity... a. from t = 0 to 10 b. from t = 10 to 15 c. from t = 15 to 40 d. from t = 40 to 55

Explanation:

Step1: Recall velocity - displacement relation

Velocity $v=\frac{\Delta x}{\Delta t}$, which is the slope of the displacement - time graph.

Step2: Analyze first graph

The displacement - time graph is a straight line. The slope is constant. So the velocity is constant and acceleration $a = 0$ (since $a=\frac{\Delta v}{\Delta t}$ and $\Delta v = 0$).

Step3: Analyze second graph for part a

For $t = 0$ to $t = 10$ s, $x_1=0$ m, $x_2 = 60$ m, $t_1 = 0$ s, $t_2=10$ s. Then $v=\frac{x_2 - x_1}{t_2 - t_1}=\frac{60 - 0}{10-0}=6$ m/s.

Step4: Analyze second graph for part b

For $t = 10$ to $t = 15$ s, the position is constant ($x = 60$ m). So $v=\frac{\Delta x}{\Delta t}=0$ m/s.

Step5: Analyze second graph for part c

For $t = 15$ to $t = 40$ s, $x_1 = 60$ m, $x_2=-40$ m, $t_1 = 15$ s, $t_2 = 40$ s. Then $v=\frac{x_2 - x_1}{t_2 - t_1}=\frac{-40 - 60}{40 - 15}=\frac{-100}{25}=-4$ m/s.

Step6: Analyze second graph for part d

For $t = 40$ to $t = 55$ s, $x_1=-40$ m, $x_2 = 0$ m, $t_1 = 40$ s, $t_2 = 55$ s. Then $v=\frac{x_2 - x_1}{t_2 - t_1}=\frac{0+40}{55 - 40}=\frac{40}{15}=\frac{8}{3}\approx2.67$ m/s.

Answer:

1. Velocity is constant, acceleration is 0.

2.
a. 6 m/s
b. 0 m/s
c. - 4 m/s
d. $\frac{8}{3}$ m/s (or approximately 2.67 m/s)