the following function describes the velocity of a car (in mi/hr) moving along a straight highway for a 4…

the following function describes the velocity of a car (in mi/hr) moving along a straight highway for a 4 - hr interval. find the function that gives the displacement of the car over the interval 0,t, where 0≤t≤4. v(t)= { 30 if 0≤t≤1.5; 40 if 1.5<t≤2; 36 if 2<t≤4 } complete the piecewise displacement function below. s(t)= { □ if 0≤t≤1.5; □ if 1.5<t≤2; □ if 2<t≤4 }

the following function describes the velocity of a car (in mi/hr) moving along a straight highway for a 4 - hr interval. find the function that gives the displacement of the car over the interval 0,t, where 0≤t≤4. v(t)= { 30 if 0≤t≤1.5; 40 if 1.5<t≤2; 36 if 2<t≤4 } complete the piecewise displacement function below. s(t)= { □ if 0≤t≤1.5; □ if 1.5<t≤2; □ if 2<t≤4 }

Answer

Explanation:

Step1: Recall displacement - velocity relation

Displacement $s(t)=\int_{0}^{t}v(u)du$.

Step2: Case 1: When $0\leq t\leq1.5$

$s(t)=\int_{0}^{t}30du$. Using the integral formula $\int kdu=ku + C$ ($k = 30$ here), we have $s(t)=30t$.

Step3: Case 2: When $1.5\lt t\leq2$

$s(t)=\int_{0}^{1.5}30du+\int_{1.5}^{t}40du$. First integral: $\int_{0}^{1.5}30du=30\times1.5 = 45$. Second integral: $\int_{1.5}^{t}40du=40(t - 1.5)$. So $s(t)=45+40(t - 1.5)=40t - 15$.

Step4: Case 3: When $2\lt t\leq4$

$s(t)=\int_{0}^{1.5}30du+\int_{1.5}^{2}40du+\int_{2}^{t}36du$.

  • $\int_{0}^{1.5}30du = 45$.
  • $\int_{1.5}^{2}40du=40\times(2 - 1.5)=20$.
  • $\int_{2}^{t}36du=36(t - 2)$. Then $s(t)=45 + 20+36(t - 2)=36t - 9$.

Answer:

[s(t)=\begin{cases}30t&\text{if }0\leq t\leq1.5\40t - 15&\text{if }1.5\lt t\leq2\36t - 9&\text{if }2\lt t\leq4\end{cases}]