Answer to Question #169138 in Physics for Maria

Question

An object moves along the x axis according to the equation 𝑥 = 3.00𝑡

2 − 2.00𝑡 + 3.00,

where x is in meters and t is in seconds. Determine (a) the average speed between t = 2.00 s

and t = 3.00 s, (b) the instantaneous speed at t = 2.00 s and at t = 3.00 s, (c) the average

acceleration between t = 2.00 s and t = 3.00 s, and (d) the instantaneous acceleration at t =

2.00 s and t = 3.00 s. (e) At what time is the object at rest?

Answer 1

(a) Let's first find the position of the object at $t=2\ s$ and $t=3\ s$:

$$x_1(t=2\ s)=3.0\cdot(2)^2-2.0\cdot2+3.0=11\ m,$$ $$x_2(t=3\ s)=3.0\cdot(3)^2-2.0\cdot3+3.0=24\ m.$$

By the definition of the average speed, we have:

$$v_{av}=\dfrac{x_2-x_1}{t_2-t_1}=\dfrac{24\ m-11\ m}{3\ s-2\ s}=13\ \dfrac{m}{s}.$$

(b) The speed is the derivative of the position of the object with respect to time:

$$v=\dfrac{dx}{dt}=\dfrac{d}{dt}(3t^2-2t+3)=6t-2.$$

Let's find the instantaneous speed at $t=2\ s$:

$$v_1(t=2\ s)=6\cdot2-2=10\ \dfrac{m}{s}.$$

Let's find the instantaneous speed at $t=3\ s$:

$$v_2(t=3\ s)=6\cdot3-2=16\ \dfrac{m}{s}.$$

(c) By the definition of the average acceleration, we have:

$$a_{avg}=\dfrac{v_2-v_1}{t_2-t_1}=\dfrac{16\ \dfrac{m}{s}-10\ \dfrac{m}{s}}{3\ s-2\ s}=6\ \dfrac{m}{s^2}.$$

(d) The acceleration is the derivative of the speed of the object with respect to time:

$$a=\dfrac{dv}{dt}=\dfrac{d}{dt}(6t-2)=6\ \dfrac{m}{s^2}.$$

As we can see from the calculations, at $t=2\ s$ and at $t=3\ s$ the acceleration remains constant and equals $6\ \dfrac{m}{s^2}$.

(e) The object is at rest when $v=0$ :

$$6t-2=0,$$ $$t=\dfrac{2}{6}=0.33\ s.$$