Question

4. A car's position as a function of time is given by the following equation: x(t)-5 m/s t+2.8 m/s2 t-0.15 m/s3 t3. a. Find the average velocity from 0 to 5 s b. Find the instantaneous velocity at 0, 3, and 5s. c. Find the average acceleration from 0 to 5 s. d. Find the instantaneous acceleration at 0, 3, and 5 s. e. At what POSITIVE time does the car come to rest?

Answer 1

$\\4)\\ a)\\ at\;time\;t_1=0\;s\\ _1=5(0)+2.8(0^2)-0.15(0^3)=0\\\\ at\;time\;t_2=5\;s\\ _2=5(5)+2.8(5^2)-0.15(5^3)=76.25\;m\\\\ average\;velocity\;v_{av}=\frac{x_2-_1}{t_2-t_1}=\frac{76.25-0}{5-0}=15.25\;m/s\\\\ (b)\\ v_t=\frac{\mathrm{d} x}{\mathrm{d} t}=5+5.6\;t-0.45\;t^2\\ at\;t=0\;\;,v=5\;m/s\\\\ at\;t=3\;\;,v=5+5.6(3)-0.45(3^2)=17.75\;m/s\\\\ at\;t=5\;\;,v=5+5.6(5)-0.45(5^2)=21.75\;m/s\\\\ (c)\\\\ average\;acceleration\;,a_{av}=\frac{v_5-v_0}{5-0}=\frac{21.75-5}{5-0}=3.35\;m/s^2\\\\ (d)\\\\ instantaneous\;acceleration\;a_t=\frac{\mathrm{d} v}{\mathrm{d} t}=5.6-0.9t\\\\ at\;t=0\;\;,a=5.6\;m/s^2\\\\ at\;t=3\;\;,a=5.6-0.9(3)=2.9\;m/s^2\\\\ at\;t=5\;\;,a=5.6-0.9(5)=1.1\;m/s^2\\\\$

$\\(e)\\if\;v_t=0\\5+5.6t-0.45t^2=0\\\\ t=13.3\;s$