Question

If velocity of a particle depends on time as v = $t^3+\frac{2}{\sqrt{t}}$. Then with what velocity is the particle moving at t = 4 sec and what is the displacement till then?

• A. v = 64, d = 64
• B. v = 65, d = 64
• C. v = 65, d = 72
• D. $v=65,d=\frac{383}{8}$

Answer 1

The correct option is C

v = 65, d = 72

Given v = $t^3+\frac{2}{\sqrt{t}}$

So at t = 4 sec

$v=(4{)}^3+\frac{2}{\sqrt{4}}=65m/s$

Now for displacement

$v=\frac{dx}{dt}=t^3+\frac{2}{\sqrt{t}}$

$\underset{0}{\overset{x}{\int }}dx$ $=\underset{0}{\overset{4}{\int }}(t^3+\frac{2}{\sqrt{t}})dt$

$[x{]}_0^x=[\frac{t^4}{4}+\frac{2}{\frac{1}{2}}(t{)}^{-\frac{1}{2}+1}{]}_0^4$

$x-0=[\frac{t^4}{4}{]}_0^4+[4\sqrt{t}{]}_0^4$

x = 64 + 8 = 72m.