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) :-
1) increases
2) decreases
3) stay constant

Answer 1

Answer:

3

Explanation:

acceleration can be expressed as a= v. (dv/dx)

Answer 2

Answer:

The magintude of the acceleration remains Cosntant and is equal to $\frac{b}{2}$

Explanation:

Acceleration-;The rate of change of velocity with respect to time is known as Acceleration

Here A=Acceleration,v=Velocity and x=Displacement

Given equation is $v^{2}=a+bx$

Differentiating with respect to x we get

$2v\frac{\mathrm{d} v}{\mathrm{d} x}=0+b$

$v\frac{\mathrm{d} v}{\mathrm{d} x}=\frac{b}{2}$

$v\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}=\frac{b}{2}$

$\textrm{As we Know}\frac{\mathrm{d} v}{\mathrm{d} t}=A\ and\ \frac{\mathrm{d} x}{\mathrm{d} t}=v$

$v.A.\frac{1}{v}=\frac{b}{2}$

$A=\frac{b}{2}$