a particle moves in a straight line with velocity v(t) meters per second (graphed), where t is time in…

a particle moves in a straight line with velocity v(t) meters per second (graphed), where t is time in seconds. at t = 1, the particles distance from the starting point was 2 meters in the positive direction. what is the particles displacement between t = 1 and t = 6 seconds? which expression should madelyn use to solve the problem? choose 1 answer: a ∫₁⁶|v(t)|dt b 2 + ∫₁⁶|v(t)|dt c 2 + ∫₁⁶v(t)dt d ∫₁⁶v(t)dt

a particle moves in a straight line with velocity v(t) meters per second (graphed), where t is time in seconds. at t = 1, the particles distance from the starting point was 2 meters in the positive direction. what is the particles displacement between t = 1 and t = 6 seconds? which expression should madelyn use to solve the problem? choose 1 answer: a ∫₁⁶|v(t)|dt b 2 + ∫₁⁶|v(t)|dt c 2 + ∫₁⁶v(t)dt d ∫₁⁶v(t)dt

Answer

Explanation:

Step1: Recall displacement formula

Displacement from $t = a$ to $t = b$ is given by $\int_{a}^{b}v(t)dt$. Here $a = 1$ and $b=6$.

Step2: Consider initial - position

The initial position at $t = 1$ is 2 meters, but displacement only depends on the integral of velocity over the time - interval and not on the initial position.

Answer:

D. $\int_{1}^{6}v(t)dt$