Question

A particle is moving in straight line. The velocity v of the particle varies with time t as $v=t^2-4t$. then the distance traveled by the particle in $t=0$ to $t=6s$ (where $t$ in second and v is in m/s )

• A. $\frac{64}{3}m$
• B. $zero$
• C. $\frac{32}{3}m$
• D. $none$

Answer 1

Answer: (B)

$zero$

Solution

we have

$V=t^2-4t$

$\Rightarrow \frac{dx}{dt}=t^2-4t$

$\Rightarrow {\int }_0^{△x}dx=\int (t^2-4t).dt$

$\Rightarrow [x{]}_0^{△x}=[\frac{t^3}{3}-2t^2{]}_0^6$

$\Rightarrow △x=\frac{(6{)}^3}{3}-2(6{)}^2$

$\Rightarrow △x=0$

Option - B is correct.