<< Chapter < Page | Chapter >> Page > |
When given the acceleration function, what additional information is needed to find the velocity function and position function?
The acceleration of a particle varies with time according to the equation $a(t)=p{t}^{2}-q{t}^{3}$ . Initially, the velocity and position are zero. (a) What is the velocity as a function of time? (b) What is the position as a function of time?
Between t = 0 and t = t _{0} , a rocket moves straight upward with an acceleration given by $a(t)=A-B{t}^{1\phantom{\rule{0.2em}{0ex}}\text{/}2}$ , where A and B are constants. (a) If x is in meters and t is in seconds, what are the units of A and B ? (b) If the rocket starts from rest, how does the velocity vary between t = 0 and t = t _{0} ? (c) If its initial position is zero, what is the rocket’s position as a function of time during this same time interval?
a.
$A={\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}B={\text{m/s}}^{5\phantom{\rule{0.2em}{0ex}}\text{/}2}$ ;
b.
$\begin{array}{}\\ \\ v(t)={\displaystyle \int a(t)dt+{C}_{1}}={\displaystyle \int \left(A-B{t}^{1\phantom{\rule{0.2em}{0ex}}\text{/}2}\right)dt+{C}_{1}}=At-\frac{2}{3}B{t}^{3\phantom{\rule{0.2em}{0ex}}\text{/}2}+{C}_{1}\hfill \\ v(0)=0={C}_{1}\phantom{\rule{0.5em}{0ex}}\text{so}\phantom{\rule{0.5em}{0ex}}v({t}_{0})=A{t}_{0}-\frac{2}{3}B{t}_{0}^{\text{3/2}}\hfill \end{array}$ ;
c.
$\begin{array}{}\\ \\ x(t)={\displaystyle \int v(t)dt+{C}_{2}={\displaystyle \int \left(At-\frac{2}{3}B{t}^{3\phantom{\rule{0.2em}{0ex}}\text{/}2}\right)dt+{C}_{2}}}=\frac{1}{2}A{t}^{2}-\frac{4}{15}B{t}^{5\phantom{\rule{0.2em}{0ex}}\text{/}2}+{C}_{2}\hfill \\ x(0)=0={C}_{2}\phantom{\rule{0.5em}{0ex}}\text{so}\phantom{\rule{0.5em}{0ex}}x({t}_{0})=\frac{1}{2}A{t}_{0}^{2}-\frac{4}{15}B{t}_{0}^{\text{5/2}}\hfill \end{array}$
The velocity of a particle moving along the x- axis varies with time according to $v(t)=A+B{t}^{\mathrm{-1}}$ , where A = 2 m/s, B = 0.25 m, and $1.0\phantom{\rule{0.2em}{0ex}}\text{s}\le t\le 8.0\phantom{\rule{0.2em}{0ex}}\text{s}$ . Determine the acceleration and position of the particle at t = 2.0 s and t = 5.0 s. Assume that $x(t=1\phantom{\rule{0.2em}{0ex}}\text{s})=0$ .
A particle at rest leaves the origin with its velocity increasing with time according to v ( t ) = 3.2 t m/s. At 5.0 s, the particle’s velocity starts decreasing according to [16.0 – 1.5( t – 5.0)] m/s. This decrease continues until t = 11.0 s, after which the particle’s velocity remains constant at 7.0 m/s. (a) What is the acceleration of the particle as a function of time? (b) What is the position of the particle at t = 2.0 s, t = 7.0 s, and t = 12.0 s?
a.
$\begin{array}{}\\ \\ a(t)=3.2{\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}t\le 5.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ a(t)=1.5{\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}5.0\phantom{\rule{0.2em}{0ex}}\text{s}\le t\le 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ a(t)=0{\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}t>11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \end{array}$ ;
b.
$\begin{array}{}\\ \\ x(t)={\displaystyle \int v(t)dt+{C}_{2}={\displaystyle \int 3.2tdt+{C}_{2}=1.6{t}^{2}+{C}_{2}}}\hfill \\ \phantom{\rule{1em}{0ex}}t\le 5.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ x(0)=0\Rightarrow {C}_{2}=0\phantom{\rule{0.5em}{0ex}}\text{therefore,}\phantom{\rule{0.2em}{0ex}}x(2.0\phantom{\rule{0.2em}{0ex}}\text{s})=6.4\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ x(t)={\displaystyle \int v(t)dt+{C}_{2}={\displaystyle \int \left[16.0-1.5\left(t-5.0\right)\right]}dt+{C}_{2}}=16t-1.5\left(\frac{{t}^{2}}{2}-5.0t\right)+{C}_{2}\hfill \\ 5.0\le t\le 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ x(5\phantom{\rule{0.2em}{0ex}}\text{s})=1.6{(5.0)}^{2}=40\phantom{\rule{0.2em}{0ex}}\text{m}=16(5.0\phantom{\rule{0.2em}{0ex}}\text{s})-1.5\left(\frac{{5}^{2}}{2}-5.0\left(5.0\right)\right)+{C}_{2}\hfill \\ \phantom{\rule{1em}{0ex}}40=98.75+{C}_{2}\Rightarrow {C}_{2}=\mathrm{-58.75}\hfill \\ x(7.0\phantom{\rule{0.2em}{0ex}}\text{s})=16(7.0)-1.5\left(\frac{{7}^{2}}{2}-5.0\left(7\right)\right)-58.75=69\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ \phantom{\rule{1em}{0ex}}x(t)={\displaystyle \int 7.0dt+{C}_{2}=7t+{C}_{2}}\hfill \\ \phantom{\rule{1.5em}{0ex}}t\ge 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ x(11.0\phantom{\rule{0.2em}{0ex}}\text{s})=16(11)-1.5\left(\frac{{11}^{2}}{2}-5.0\left(11\right)\right)-58.75=109=7(11.0\phantom{\rule{0.2em}{0ex}}\text{s})+{C}_{2}\Rightarrow {C}_{2}=32\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ \phantom{\rule{1em}{0ex}}x(t)=7t+32\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ \phantom{\rule{1.5em}{0ex}}x\ge 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\Rightarrow x(12.0\phantom{\rule{0.2em}{0ex}}\text{s})=7(12)+32=116\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \end{array}$
Professional baseball player Nolan Ryan could pitch a baseball at approximately 160.0 km/h. At that average velocity, how long did it take a ball thrown by Ryan to reach home plate, which is 18.4 m from the pitcher’s mound? Compare this with the average reaction time of a human to a visual stimulus, which is 0.25 s.
An airplane leaves Chicago and makes the 3000-km trip to Los Angeles in 5.0 h. A second plane leaves Chicago one-half hour later and arrives in Los Angeles at the same time. Compare the average velocities of the two planes. Ignore the curvature of Earth and the difference in altitude between the two cities.
Take west to be the positive direction.
1st plane:
$\stackrel{\text{\u2013}}{\nu}=600\phantom{\rule{0.2em}{0ex}}\text{km/h}$
2nd plane
$\stackrel{\text{\u2013}}{\nu}=667.0\phantom{\rule{0.2em}{0ex}}\text{km/h}$
Unreasonable Results A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. If his average velocity is 24 km/h, how long did it take him to complete the trip? Is this a reasonable time?
Notification Switch
Would you like to follow the 'University physics volume 1' conversation and receive update notifications?