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Standard XIII

Physics

Uniform and Non Uniform

The speed v o...

Question

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

Increases

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Decreases

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Stays constant

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First decreases and then increases

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Solution

The correct option is C Stays constant
$v = \sqrt{a + b x}$
We know that,
$a = v \frac{d v}{d x}$
$= \sqrt{a + b x} \times \frac{b}{2 \sqrt{a + b x}}$
$= \frac{b}{2}$ (constant)
Alternate solution:
$v^{2} = a + b x$
Comparing with
$v^{2} = u^{2} + 2 a s$
$\Rightarrow 2 a = b \Rightarrow a = \frac{b}{2}$ (constant).

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The speed v of a car moving on a straight road changes according to equation, v2=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) :

The correct option is C Stays constantv=√a+bx We know that, a=vdvdx =√a+bx×b2√a+bx =b2 (constant) Alternate solution: v2=a+bx Comparing with v2=u2+2as ⇒2a=b⇒a=b2 (constant).

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