Question

The distance traveled by car in $\mathrm{t}$ seconds after applying the brakes is $\mathrm{s}$ feet, where $\mathrm{s}=22\mathrm{t}–12{\mathrm{t}}^2$. The car travels a certain distance before coming to a complete halt.

• A. $10.08ft$
• B. $10ft$
• C. $11ft$
• D. $11.5ft$

Answer 1

The correct option is A

$10.08ft$

Step: 1 Given data:

The given equation of motion is $\mathrm{s}=22\mathrm{t}–12{\mathrm{t}}^2$

where $\mathrm{s}=$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 Distance:

$\begin{array}{rcl}\mathrm{s}& =& 22\mathrm{t}–12{\mathrm{t}}^2\\ \mathrm{v}& =& \frac{\mathrm{d}\left(22\mathrm{t}-12{\mathrm{t}}^2\right)}{\mathrm{dt}}\\ & =& 22-24\mathrm{t}\end{array}$

for the time when the body will stop, the velocity will be zero, so

$\begin{array}{rcl}22-24\mathrm{t}& =& 0\\ 22& =& 24\mathrm{t}\\ \frac{22}{24}& =& \mathrm{t}\\ & & \mathrm{or}\\ \mathrm{t}& =& \frac{11}{12}\mathrm{seconds}\end{array}$

Step 4: Calculating distance

$$\begin{gathered} \mathrm{s}=22\left(\frac{11}{12}\right)–12{\left(\frac{11}{12}\right)}^2 \\ =\frac{121}{12}\mathrm{m}\mathrm{or}10.08\mathrm{ft} \end{gathered}$$

Hence option A is correct