Question

A particle starts from $x=0$ and moves along a straight line such that its velocity $v$ depends on position $x$ as $v=x+4$.
For the first four seconds, average speed of the particle (in $m/s$) is

• A. $e^4+1$
• B. $e^4-1$
• C. $e^4+4$
• D. $e^4-4$

Answer 1

The correct option is B $e^4-1$

Given, $v=x+4$
$\frac{dx}{dt}=x+4$
Integrating on both sides
$\int \frac{dx}{x+4}=\int dt$
$ln(x+4)=t+C$
$x+4=e^{(t+C)}$
At $t=0,x=0$
$\Rightarrow x=4e^t-4$
$v=4e^t>0\text{for}t>0$
$\Rightarrow \text{Total distance}=\text{Total displacement}$
∴ Average speed for first $4$ seconds $=\frac{\text{Total distance}}{\text{Total time}}$
$=\frac{4e^4-4}{4}=e^4-1$