The acceleration function (in m / s²) and the initial velocity are given for a particle moving along a line. Find the velocity at time t and the distance traveled during the given time interval.

Question

Quizplus · Accepted Answer

Using integration, we can find the velocity function from the acceleration function:

$v(t)=\int a(t) \ dt = \int 4t \ dt = 2t^2 + C$

Since the initial velocity is given as $v(0)=5$, we can find the value of C:

$v(0)=5=2(0)^2+C \Rightarrow C=5$

Thus, the velocity function is:

$v(t)=2t^2+5$

To find the distance traveled during the given time interval, we need to find the displacement function and integrate it over the given time interval:

$d(t)=\int v(t) \ dt = \int (2t^2+5) \ dt = \frac{2}{3}t^3+5t+D$

Since the particle starts at $s(0)=2$, we can find the value of D:

$s(0)=2=\frac{2}{3}(0)^3+5(0)+D \Rightarrow D=2$

Thus, the displacement function is:

$s(t)=\frac{2}{3}t^3+5t+2$

The distance traveled during the given time interval $[0,2]$ is:

$s(2)-s(0)=\frac{2}{3}(2)^3+5(2)+2-2=\frac{44}{3}$

Therefore, the correct answer is B) 11eaa3e5_559d_459e_9f8f_15dc6530d381_TB5971_11.

The Acceleration Function (In M / S2) and the Initial

The Acceleration Function (In M / S²) and the Initial