Question

The differential equation of a body in motion is given by $\frac{dv}{dt}=k(1-\frac{t}{T})$.where $k$ and $T$ are constant.Determine the distance travelled at max speed

• A. $\frac{kt}{2}$
• B. $\frac{k{t}^2}{2}$
• C. $\frac{k{t}^2}{3}$
• D. $\frac{k{t}^2}{4}$

Answer 1

Answer: (B)

$\frac{k{t}^2}{2}$

Given Differential equation is $\frac{dv}{dt}=k(1-\frac{t}{T})$

At $v_{max}$ the value of $\frac{dv}{dt}=0$,

$⟹t=Tsec$ $v_{max}$ occurs,

Now solving the differential equation gives,

Assuming that the body had started from rest,

${\int }_0^{v_{max}}dv={\int }_0^{T}k(1-\frac{t}{T})dt$,

$\therefore v_{max}=\frac{kt}{2}$,where $t=Tunits$,

$\therefore$the disance travelled till $v_{max}$ occurs is

${\int }_0^T(kt-\frac{t^2}{2T})dt$,

$⟹\frac{k{t}^2}{2}$ where $t=Tunits$