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A particle moves along the 
x-axis such that at any time 
t >= 0 its position is 
x(t), its velocity is 
v(t), and its acceleration is 
a(t).
What is the average velocity of the particle on the interval 
1 <= t <= 8 ?

(v(8)-v(1))/(7)

(a(8)-a(1))/(7)

(1)/(7)int_(1)^(8)a(t)dt

(1)/(7)int_(1)^(8)v(t)dt

A particle moves along the x x -axis such that at any time t0 t \geq 0 its position is x(t) x(t) , its velocity is v(t) v(t) , and its acceleration is a(t) a(t) .\newlineWhat is the average velocity of the particle on the interval 1t8 1 \leq t \leq 8 ?\newlinev(8)v(1)7 \frac{v(8)-v(1)}{7} \newlinea(8)a(1)7 \frac{a(8)-a(1)}{7} \newline1718a(t)dt \frac{1}{7} \int_{1}^{8} a(t) d t \newline1718v(t)dt \frac{1}{7} \int_{1}^{8} v(t) d t

Full solution

Q. A particle moves along the x x -axis such that at any time t0 t \geq 0 its position is x(t) x(t) , its velocity is v(t) v(t) , and its acceleration is a(t) a(t) .\newlineWhat is the average velocity of the particle on the interval 1t8 1 \leq t \leq 8 ?\newlinev(8)v(1)7 \frac{v(8)-v(1)}{7} \newlinea(8)a(1)7 \frac{a(8)-a(1)}{7} \newline1718a(t)dt \frac{1}{7} \int_{1}^{8} a(t) d t \newline1718v(t)dt \frac{1}{7} \int_{1}^{8} v(t) d t
  1. Define average velocity: Understand the definition of average velocity.\newlineAverage velocity is defined as the change in position over the change in time. Since we are given the velocity function v(t)v(t), we can find the change in position by integrating v(t)v(t) over the interval from t=1t=1 to t=8t=8.
  2. Set up integral: Set up the integral to find the change in position.\newlineThe change in position, Δx\Delta x, from t=1t=1 to t=8t=8 is given by the integral of the velocity function over that interval:\newlineΔx=t=1t=8v(t)dt\Delta x = \int_{t=1}^{t=8} v(t) \, dt
  3. Calculate average velocity: Calculate the average velocity.\newlineThe average velocity, vavgv_{\text{avg}}, is the change in position, Δx\Delta x, divided by the change in time, Δt\Delta t. Since the interval is from t=1t=1 to t=8t=8, Δt\Delta t is 81=78 - 1 = 7. Therefore, the average velocity is:\newlinevavg=17×t=1t=8v(t)dtv_{\text{avg}} = \frac{1}{7} \times \int_{t=1}^{t=8} v(t) \, dt
  4. Identify correct expression: Identify the correct expression for the average velocity.\newlineThe correct expression for the average velocity is the one we derived in Step 33, which is:\newlinevavg=17×t=1t=8v(t)dtv_{\text{avg}} = \frac{1}{7} \times \int_{t=1}^{t=8} v(t) \, dt\newlineThe other options provided are incorrect because they involve the acceleration function a(t)a(t) or the velocities at specific points in time, which are not directly related to the average velocity over an interval.

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