Question

The speed $v$ of a car moving on a straight road changes according to equation, $v^2=a+bx$, where $a$ and $b$ are positive constants. Then, the magnitude of acceleration in the course of such motion: ($x$ is the distance travelled) :

• A. Increases
• B. Decreases
• C. Stays constant
• D. First decreases and then increases

Answer 1

The correct option is C Stays constant

$v=\sqrt{a+bx}$
We know that,
$a=v\frac{dv}{dx}$
$=\sqrt{a+bx}\times \frac{b}{2\sqrt{a+bx}}$
$=\frac{b}{2}$ (constant)
Alternate solution:
$v^2=a+bx$
Comparing with
$v^2=u^2+2as$
$\Rightarrow 2a=b\Rightarrow a=\frac{b}{2}$ (constant).