Question

I have been tryinng to figure this out: An object moves with constant acceleration 3.65 m/s2 and over a time interval reaches a final velocity of 11.4 m/s.
(a) If its initial velocity is 5.7 m/s, what is its displacement during the time interval?

(b) What is the distance it travels during this interval?

(c) If its initial velocity is -5.7 m/s, what is its displacement during the time interval?

(d) What is the total distance it travels during the interval in part (c)?

Answer 1

Condition:

I have been trying to figure this out: An object moves with constant acceleration $3.65 \, \mathrm{m/s^2}$ and over a time interval reaches a final velocity of $11.4 \, \mathrm{m/s}$.

(a) If its initial velocity is $5.7 \, \mathrm{m/s}$, what is its displacement during the time interval?

(b) What is the distance it travels during this interval?

(c) If its initial velocity is $-5.7 \, \mathrm{m/s}$, what is its displacement during the time interval?

(d) What is the total distance it travels during the interval in part (c)?

Solution

Assume that positive $x$-axis directed to the right

(a)(b)

$$v = v_{00} + at$$ $$t = \frac{(v - v_{00})}{a} = \frac{(11.4 - 5.7)}{3.65} = 1.56s$$

displacement = distance $= v_{00}t + \frac{at^2}{2} = 5.7 * 1.56 + \frac{3.65 * 1.56^2}{2} = 8.892 + 4.44 = 13.332 \, m$

(from origin point - to theright)

(c)

$$v = -v_{00} + at$$ $$t_1 = \frac{(0 + v_{00})}{a} = \frac{5.7}{3.65} = 1.56 \, s.$$ $$s_1 = \frac{v_0^2}{2a} = \frac{5.7^2}{2 * 3.65} = 4.5 \, m \quad (\text{to the left})$$ $$v = a * t_2$$ $$t_2 = \frac{v}{a} = \frac{11.4}{3.65} = 3.12 \, s.$$ $$s_2 = \frac{at_2^2}{2} = 3.65 * \frac{3.12^2}{2} = 17.77 \, m \quad (\text{to the right})$$ $$\text{Displacement} = 17.77 - 4.5 = 13.27 \, m$$

(d)

$$\text{Distance} = 4.5 + 17.77 = 22.27 \, m$$

Answers: a,b: displacement = distance $= 13.332 \, m$; c: Displacement = $13.27 \, m$; d: Distance = $22.27 \, m$.