Question

The acceleration- time relationship for a vehicle subjected to non- uniform acceleration is

$\frac{dv}{dt}=(\alpha -\beta v_0)e^{-\beta t}$

Where, v is the speed in m/s, t is the time in s, $\alpha$ and $\beta$ are parameters, and $v_0$ is the initial speed in m/s. If the accelerating behaviour of a vehicle, whose driver intends to overtake a slow moving vehicle ahead, is described as,

$\frac{dv}{dt}=(\alpha -\beta v)$

Considering
$\alpha =2m/s^2,\beta =0.05s^{-1}\text{and}$

$\frac{dv}{dt}=1.3m/s^{2}att=3s$, the distance (in m) travelled by the vehicle in 35 sec is

• A. 900.96

Answer 1

The correct option is A 900.96
$\frac{dv}{dt}=\alpha -\beta v$

$\Rightarrow \int \frac{dv}{(\alpha -\beta v)}=\int dt$

$\Rightarrow \frac{ln(\alpha -\beta v)}{-\beta }=t+C$

at t = 3 sec,
$\frac{dv}{dt}=\alpha -\beta v=1.3m/s^2$

$\Rightarrow \frac{ln\left(1.3\right)}{-0.05}=3+c\Rightarrow c=-8.247$

$\therefore ln(2-0.05\times v)=(t-8.247)\times (-0.05)$

$\Rightarrow 2-0.05\times v=1.51e^{-0.05t}$

$\Rightarrow v=\frac{ds}{dt}=40-30.2e^{-0.05t}$

$\Rightarrow s=40t+604e^{-0.05t}+c^{\prime }$

$\therefore att=0,s=0,$

$\Rightarrow c^{\prime }=-604$

$\Rightarrow s=40t+604e^{-0.05t}-604$

at t = 35 sec

$\Rightarrow s=40\times 35+604\times e^{-0.05\times 35}-604$

$\Rightarrow s=900.96m$