Question

A particle's motion is given by $s=a(t+b(e^{(-\frac{t}{b})}-1))$, where ($s$ is in metre, $a$ and $b$ are constants and $t$ is in sec). What will be the instantaneous speed of the particle at $t=\infty$

• A. $a$
• B. $b$
• C. $\infty$
• D. $0$

Answer 1

The correct option is A $a$

As we know that instantneous speed $=\frac{ds}{dt}=\frac{d\left(a(t+b(e^{-\frac{t}{b}}-1))\right)}{dt}\Rightarrow v=a\frac{dt}{dt}+ab\frac{d(e^{-\frac{t}{b}}-1)}{dt}\Rightarrow v=a(1-e^{\frac{-t}{b}})$

Putting the value of $t=\infty$, in the equation we get $v=a$