Question

If the velocity of a particle is given by $v=(180-16x{)}^{\frac{1}{2}}m/s$ then its acceleration will be

• A. $Zero$
• B. $8m/s^2$
• C. $-8m/s^2$
• D. $4m/s^2$

Answer 1

The correct option is C $-8m/s^2$

$v=(180-16x{)}^{\frac{1}{2}}$

As $a=\frac{dv}{dt}=\frac{dv}{dx}.\frac{dx}{dt}$

$\therefore a=\frac{1}{2}(180-16x{)}^{\frac{-1}{2}}\times (-16\left)\right(\frac{dx}{dt})$

$=-8(180-16x{)}^{\frac{-1}{2}}\times v$

$=-8(180-16x{)}^{\frac{-1}{2}}\times (180-16x{)}^{\frac{1}{2}}=-8m/s^2$