Question

When a particle's distance traveled is proportional to the squares of time, the particle travels with

• A. Uniform acceleration
• B. Uniform velocity
• C. Increasing acceleration
• D. Decreasing velocity

Answer 1

The correct option is A

Uniform acceleration

Step: 1 Given data:

The equation of motion is $\mathrm{s}={\mathrm{kt}}^2$

where $s=$distance, $k=$constant, and $t=$time

Step: 2 Formula used:

Acceleration$\left(\mathrm{a}\right)=\frac{{\mathrm{d}}^2\mathrm{s}}{{\mathrm{dt}}^2}$

Step: 3 Calculate Acceleration:

$\begin{array}{rcl}\mathrm{a}& =& \frac{{\mathrm{d}}^2\mathrm{s}}{{\mathrm{dt}}^2}\\ & =& \frac{{\mathrm{d}}^2\left({\mathrm{kt}}^2\right)}{\mathrm{dt}}\\ & =& 2\mathrm{k}\\ 2\mathrm{k}& =& \mathrm{constant}\end{array}$

Therefore the acceleration will be uniform

Hence option A is correct.