Question

The velocity function, in meters per minute, of a particle moving along a straight line isv(t)=5t−2,where t is time in minutes.1. Find the displacement of the particle from t=2 minutes to t=5 minutes?meters Tries 0/992. Find the time t when the displacement is zero after the particle starts moving?minutes

Answer 1

Part A. We are given that the velocity function of a particle is given by:

$v(t)=5t-2$

We are asked to determine the displacement. To do that we need to determine the formula for the distance. Let's remember that the distance is given by:

$s(t)=\int v(t)dt$

Now, we substitute the formula for the velocity:

$s(t)=\int(5t-2)dt$

Now, we distribute the integral:

$s(t)=\int5tdt-\int2dt$

Now, we use the following formula for integration:

$\int x^ndx=\frac{x^{n+1}}{n+1}+c$

Applying the rule we get:

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

Now, to determine the displacement we need to determine the difference between the positions, like this:

$d=s(5)-s(2)$

Now, we substitute the values in the formula for the position:

$\begin{gathered} s(5)=\frac{5(5)^2}{2}-2(5)+c=\frac{105}{2}+c \\ \\ s(2)=\frac{5(2^2)}{2}-2(2)+c=6+c \end{gathered}$

Now, we substitute in the formula for the displacement:

$d=\frac{105}{2}+c-6-c$

Solving the operations:

$d=\frac{93}{2}=46.5$

Therefore, the displacement is 46.5 minutes

Part B. The displacement of the particle between an initial time "t = 0" and a final time "t" is:

$d=s(t)-s(0)$

Substituting we get:

$d=\frac{5t^2}{2}-2t+c-\frac{5(0)^2}{2}+2(0)-c$

Solving the operations:

$d=\frac{5t^2}{2}-2t$

Now, we set the displacement to zero:

$\frac{5t^2}{2}-2t=0$

Now, we take "t" as a common factor:

$t(\frac{5t}{2}-2)=0$

Since we want the value of time that is different from zero we set the second factor to zero:

$\frac{5t}{2}-2=0$

Now, we add 2 to both sides:

$\frac{5t}{2}=2$

Now, we multiply both sides by 2:

$5t=4$

Now, we divide both sides by 5:

$t=\frac{4}{5}=0.8$

Therefore, the time is 0.8 minutes.