Question

10

Answer 1

Answer:

Acceleration when time is $2 \ sec$ = $-\frac{5}{4} \ m/s^2$

Distance $s=9-\frac{9}{t}$

Distance travelled between $t=1$ and $t=3$ is $6\ m$

Step-by-step explanation:

Velocity $V=1+\frac{9}{t^2} \ \ \ when \ \ 1\leq t\leq 3$

Acceleration(a): Rate of change of velocity.

$a=\frac{dV}{dt} =1-\frac{18}{t^3}$

at $t=2$

$a=1-\frac{18}{8} \\=1-\frac{9}{4} \\=-\frac{5}{4}$

$a=-\frac{5}{4} \ m/s^2$

Distance(s):

$V=\frac{ds}{dt}\\\\\frac{ds}{dt}=1+\frac{9}{t^2}\\\\Integrate\ both\ sides\\\\s=1-\frac{9}{t}+k\\\\ k\ is\ a\ constant\\\\Given\ s=6\ when\ t=3\\\\6=1-\frac{9}{3}+k\\\\ k=8\\\\Hence\ s=1-\frac{9}{t}+8\\\\ s=9-\frac{9}{t}$

Distance between $t=1 \ and \ t=3$

$=s(3)-s(1)\\\\=(9-\frac{9}{3})-(9-9) \\\\=6 \ m$