3.6 Finding velocity and displacement from acceleration  (Page 3/6)

a. $A={\text{m/s}}^{2}B={\text{m/s}}^{5\text{/}2}$ ;
b. $\begin{array}{}\\ \\ v(t)={\displaystyle \int a(t)dt+C_1}={\displaystyle \int \left(A-B{t}^{1\text{/}2}\right)dt+C_1}=At-\frac{2}{3}B{t}^{3\text{/}2}+C_1\hfill \\ v(0)=0=C_1\text{so}v(t_0)=A{t}_0-\frac{2}{3}B{t}_0^{\text{3/2}}\hfill \end{array}$ ;
c. $\begin{array}{}\\ \\ x(t)={\displaystyle \int v(t)dt+C_2={\displaystyle \int \left(At-\frac{2}{3}B{t}^{3\text{/}2}\right)dt+C_2}}=\frac{1}{2}A{t}^2-\frac{4}{15}B{t}^{5\text{/}2}+C_2\hfill \\ x(0)=0=C_2\text{so}x(t_0)=\frac{1}{2}A{t}_0^2-\frac{4}{15}B{t}_0^{\text{5/2}}\hfill \end{array}$

a. $\begin{array}{}\\ \\ a(t)=3.2{\text{m/s}}^{2}t\le 5.0\text{s}\hfill \\ a(t)=1.5{\text{m/s}}^{2}5.0\text{s}\le t\le 11.0\text{s}\hfill \\ a(t)=0{\text{m/s}}^{2}t>11.0\text{s}\hfill \end{array}$ ;
b. $\begin{array}{}\\ \\ x(t)={\displaystyle \int v(t)dt+C_2={\displaystyle \int 3.2tdt+C_2=1.6t^2+C_2}}\hfill \\ t\le 5.0\text{s}\hfill \\ x(0)=0\Rightarrow C_2=0\text{therefore,}x(2.0\text{s})=6.4\text{m}\hfill \\ x(t)={\displaystyle \int v(t)dt+C_2={\displaystyle \int \left[16.0-1.5\left(t-5.0\right)\right]}dt+C_2}=16t-1.5\left(\frac{t^2}{2}-5.0t\right)+C_2\hfill \\ 5.0\le t\le 11.0\text{s}\hfill \\ x(5\text{s})=1.6{(5.0)}^2=40\text{m}=16(5.0\text{s})-1.5\left(\frac{5^2}{2}-5.0\left(5.0\right)\right)+C_2\hfill \\ 40=98.75+C_2\Rightarrow C_2=\mathrm{-58.75}\hfill \\ x(7.0\text{s})=16(7.0)-1.5\left(\frac{7^2}{2}-5.0\left(7\right)\right)-58.75=69\text{m}\hfill \\ x(t)={\displaystyle \int 7.0dt+C_2=7t+C_2}\hfill \\ t\ge 11.0\text{s}\hfill \\ x(11.0\text{s})=16(11)-1.5\left(\frac{{11}^2}{2}-5.0\left(11\right)\right)-58.75=109=7(11.0\text{s})+C_2\Rightarrow C_2=32\text{m}\hfill \\ x(t)=7t+32\text{m}\hfill \\ x\ge 11.0\text{s}\Rightarrow x(12.0\text{s})=7(12)+32=116\text{m}\hfill \end{array}$

Take west to be the positive direction.
1st plane: $\stackrel{\text{–}}{\nu }=600\text{km/h}$
2nd plane $\stackrel{\text{–}}{\nu }=667.0\text{km/h}$