kiran received the following problem: a particle moves in a straight line with velocity v(t)=2t² - 5t meters…

kiran received the following problem: a particle moves in a straight line with velocity v(t)=2t² - 5t meters per second, where t is time in seconds. at t = 3, the particles distance from the starting point was 9 meters in the positive direction. what is the particles position at t = 4 seconds? which expression should kiran use to solve the problem? choose 1 answer: a 9+∫₃⁴v(t)dt b v(4)-v(3) c 9+v(3) d ∫₀⁴v(t)dt

kiran received the following problem: a particle moves in a straight line with velocity v(t)=2t² - 5t meters per second, where t is time in seconds. at t = 3, the particles distance from the starting point was 9 meters in the positive direction. what is the particles position at t = 4 seconds? which expression should kiran use to solve the problem? choose 1 answer: a 9+∫₃⁴v(t)dt b v(4)-v(3) c 9+v(3) d ∫₀⁴v(t)dt

Answer

Answer:

A. $9+\int_{3}^{4}v(t)dt$

Explanation:

Step1: Recall position - velocity relation

The change in position of a particle over an interval $[a,b]$ is given by $\int_{a}^{b}v(t)dt$, where $v(t)$ is the velocity function.

Step2: Determine initial condition

The particle's position at $t = 3$ is 9 meters. To find the position at $t=4$, we add the change in position from $t = 3$ to $t = 4$ to the position at $t = 3$. The change in position from $t=3$ to $t = 4$ is $\int_{3}^{4}v(t)dt$. So the position at $t = 4$ is $9+\int_{3}^{4}v(t)dt$.