Question

$\mathrm{s}=3–4\mathrm{t}+5{\mathrm{t}}^2$ is the distance traveled by an item along a straight line in time $\mathrm{t}$, and the beginning velocity of the object is

• A. $3units$
• B. $–3units$
• C. $4units$
• D. $–4units$

Answer 1

The correct option is D

$–4units$

Step: 1 Given data:

The given equation of motion is, $\mathrm{s}=3–4\mathrm{t}+5{\mathrm{t}}^2$

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

Step: 2 Formula used:

The velocity of the particle $\left(\mathrm{v}\right)=\frac{\mathrm{ds}}{\mathrm{dt}}$

Step: 3 :Calculate velocity:

$\mathrm{s}=3–4\mathrm{t}+5{\mathrm{t}}^2$

$\begin{array}{rcl}\mathrm{v}& =& \frac{\mathrm{ds}}{\mathrm{dt}}\\ & =& \frac{\mathrm{d}\left(3-4\mathrm{t}+5{\mathrm{t}}^2\right)}{\mathrm{dt}}\\ & =& -4+10\mathrm{t}\end{array}$

Step 3: Calculating the initial velocity:

For initial velocity $\mathrm{t}=0$

$\begin{array}{rcl}{\mathrm{v}}_{\mathrm{initial}}& =& -4+10\left(0\right)\\ & =& -4\end{array}$

Hence option D is correct