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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). You are given:

x(1)=2quad" and "quad v(1)=5
Which of the following expressions gives the distance traveled by the particle over the interval 
1 <= t <= 6 ?

int_(1)^(6)v(t)dt

int_(1)^(6)|v(t)|dt

int_(1)^(6)x(t)dt

int_(1)^(6)|x(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) . You are given:\newlinex(1)=2 and v(1)=5 x(1)=2 \quad \text { and } \quad v(1)=5 \newlineWhich of the following expressions gives the distance traveled by the particle over the interval 1t6 1 \leq t \leq 6 ?\newline16v(t)dt \int_{1}^{6} v(t) d t \newline16v(t)dt \int_{1}^{6}|v(t)| d t \newline16x(t)dt \int_{1}^{6} x(t) d t \newline16x(t)dt \int_{1}^{6}|x(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) . You are given:\newlinex(1)=2 and v(1)=5 x(1)=2 \quad \text { and } \quad v(1)=5 \newlineWhich of the following expressions gives the distance traveled by the particle over the interval 1t6 1 \leq t \leq 6 ?\newline16v(t)dt \int_{1}^{6} v(t) d t \newline16v(t)dt \int_{1}^{6}|v(t)| d t \newline16x(t)dt \int_{1}^{6} x(t) d t \newline16x(t)dt \int_{1}^{6}|x(t)| d t
  1. Consider Velocity Function: To find the distance traveled by a particle over a time interval, we need to consider the velocity function, because distance is the integral of velocity with respect to time.
  2. Net Displacement vs Total Distance: The expression 16v(t)dt\int_{1}^{6}v(t)\,dt represents the net displacement of the particle from time t=1t=1 to t=6t=6, not the total distance traveled. If the velocity is always positive or always negative over the interval, then this expression would also give the total distance traveled.
  3. Account for Change in Direction: However, if the velocity changes sign (meaning the particle changes direction) within the interval from t=1t=1 to t=6t=6, then we need to take the absolute value of the velocity to ensure we are summing the distances traveled in each direction. This is represented by the expression 16v(t)dt\int_{1}^{6}|v(t)|\,dt.
  4. Incorrect Expressions: The expressions 16x(t)dt\int_{1}^{6}x(t)\,dt and 16x(t)dt\int_{1}^{6}|x(t)|\,dt are not correct for finding the distance traveled. The first one would give the net change in position (displacement) if x(t)x(t) were the velocity function, and the second one does not have a standard interpretation in this context.
  5. Correct Expression for Distance: Therefore, the correct expression for the distance traveled by the particle over the interval 1t61 \le t \le 6 is 16v(t)dt\int_{1}^{6}|v(t)|\,dt, as it accounts for the absolute value of the velocity, summing up all distances traveled regardless of direction.

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