Question

The distance covered by a moving body is directly proportional to the square of the time. The acceleration of the body is:

• A. increasing
• B. decreasing
• C. zero
• D. constant

Answer 1

Answer: (D)

constant

$\underset{distance}{s}\propto \underset{time}{t^2}\Rightarrow s=\underset{someconstant}{k{t}^2}$

Acceleration, $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\frac{d\left(\frac{d\overrightarrow{s}}{dt}\right)}{dt}=\frac{d^{2}s}{d{t}^2}$

Hence, $\overrightarrow{a}=\frac{d^{2}s}{d{t}^2}=\frac{d^2(k{t}^{2)}}{d{t}^2}=\frac{d\left(2kt\right)}{dt}=2k$

Therefore, $\overrightarrow{a}$ is constant $=2k$

It can be 0, but it it not necessarily 0.