suppose a particle moves along a straight line. the position at time t is given by ( s(t) = 7t - 2t^{2} ), (…

suppose a particle moves along a straight line. the position at time t is given by ( s(t) = 7t - 2t^{2} ), ( tgeq0 ), where t is measured in seconds and ( s(t) ) is measured in meters. complete parts (a) - (d) below. (iii) how far to the right on the straight line does the particle travel? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the particle goes at most meters to the right. b. the graph of ( s(t) ) has no maximum. the particle will continue indefinitely to the right.

suppose a particle moves along a straight line. the position at time t is given by ( s(t) = 7t - 2t^{2} ), ( tgeq0 ), where t is measured in seconds and ( s(t) ) is measured in meters. complete parts (a) - (d) below. (iii) how far to the right on the straight line does the particle travel? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the particle goes at most meters to the right. b. the graph of ( s(t) ) has no maximum. the particle will continue indefinitely to the right.

Answer

Explanation:

Step1: Find the derivative of (s(t))

The position function is (s(t)=7t - 2t^{2}). Using the power rule ((x^{n})^\prime=nx^{n - 1}), the derivative (s^\prime(t)=\frac{d}{dt}(7t-2t^{2})=7-4t).

Step2: Find the critical point

Set (s^\prime(t) = 0). So, (7-4t = 0). Solving for (t): [ \begin{align*} 7-4t&=0\ 4t&=7\ t&=\frac{7}{4}=1.75 \end{align*} ]

Step3: Find the second - derivative of (s(t))

Differentiate (s^\prime(t)) with respect to (t). (s^{\prime\prime}(t)=\frac{d}{dt}(7 - 4t)=-4). Since (s^{\prime\prime}(1.75)=-4<0), the function (s(t)) has a maximum at (t = 1.75).

Step4: Find the maximum value of (s(t))

Substitute (t = 1.75) into (s(t)): [ \begin{align*} s(1.75)&=7\times1.75-2\times(1.75)^{2}\ &=12.25-2\times3.0625\ &=12.25 - 6.125\ &=6.125 \end{align*} ]

Answer:

A. The particle goes at most (6.125) meters to the right.