Question

The velocity of a particle is at any time related to the distance travelled by the particle by the relation $v=ax+b$, where a is positive and b $\le \frac{a}{2}.$ Which of the following statements will be true for this motion? (given x = 0 when t = 0)

• A. The displacement of the particle at time t is $x=\frac{b}{a}(e^{at}-1)$
• B. The particle will experience a retardation if b < 0
• C. The particle will be at rest at time t = 0
• D. The motion of the particle is under constant acceleration.

Answer 1

Answer: (A), (B)

The correct options are
A The displacement of the particle at time t is $x=\frac{b}{a}(e^{at}-1)$
B The particle will experience a retardation if b < 0

$v=ax+b;wherea\to +veandb\le \frac{a}{2}$
$\frac{dx}{dt}=ax+b\Rightarrow \int \frac{dx}{ax+b}=\int dt$
$\Rightarrow \frac{ln(ax+b)}{a}=t+c\left(1\right)$
given at t = 0, x = 0,
$\frac{ln\left(b\right)}{a}=cputin\left(1\right)$
$\therefore \frac{ln(ax+b)}{a}=t+\frac{ln\left(b\right)}{a}$
$\Rightarrow \frac{1}{a}ln\left(\frac{ax+b}{b}\right)=t$
$\Rightarrow \frac{ax+b}{b}=e^{at}$
$\Rightarrow x=\frac{b}{a}(e^{at}-1)$
and, $v=ax+b=a\left(\frac{b}{a}\right(e^{at}-1\left)\right)+b=b{e}^{at}$
$acceleration=ab{e}^{at}$
$\therefore Forb<0,$ acceleration will be negative i.e., particle will experience retardation
$v=ax+b=b{e}^{at}$
at t = 0, v = b i.e., not at rest.
acceleration $=ab{e}^{at}$
acceleration is varying with time