Question

8 Calculate the Velocity and accleration
of a particle if the postion of the particle
relative to orgin is given as
x = 7tcube +8 t² +5
and t = 5 sec​

Answer 1

$\color{red}\huge{\underline{\underline{\bf GIVEN\dag}}}$

• Displacement as a function of time.
• $7 {t}^{3} + 8 {t}^{2} + 5$

$\color{red}\huge{\underline{\underline{\bf SOLUTION\dag}}}$

At t=5 sec displacement;

$\implies x = 7 {(5)}^{3} + 8 {(5)}^{2} + 5$

$\implies x = 7(125) + 8(25) + 5$

$\implies x = 875 + 200 + 5$

$\therefore displacement(x) = 1080$

We know that;

$\blue{ \boxed{velocity(v) = \frac{displacement}{time}} }$

$\implies v = \frac{1080}{5}$

$\orange{ \boxed{\therefore velocity(v) = 216 \: {ms}^{ - 1} }}$

HOPE THAT HELPS!!