Question

After the brakes are engaged, the distance traveled by bus in $t$ seconds equals $1+2t-2t^2$ meters. The bus travels a distance equivalent to before stopping is

• A. $0.5m$
• B. $1m$
• C. $1.5m$
• D. $2.5m$

Answer 1

The correct option is C

$1.5m$

Step: 1 Given data

Traveled distance is $s=1+2t-2t^2\dots \dots \dots \dots \left(1\right)$

Step: 2 Formula used

The relation between velocity and displacement is,

$v=\frac{\mathrm{ds}}{\mathrm{dt}}$, where $v=$velocity, $s=$displacement, and $t=$time.

Step: 3 Calculate the time when the bus stop moving:

To obtain the velocity, differentiate the equation $\left(1\right)$.

$\begin{array}{rcl}\frac{ds}{dt}& =& 0+2-4t\\ v& =& 2-4t\end{array}$

For the time when the bus stops moving, velocity $=0$

$\begin{array}{rcl}2-4t& =& 0\\ -4t& =& -2\\ t& =& \frac{1}{2}s\end{array}$

Step: 4 Calculate the distance travelled by the bus before stopping:

So, the distance traveled in $\frac{1}{2}s$ is,

$$\begin{gathered} s=1+2t-2t^2 \\ s=1+2\left(\frac{1}{2}\right)-2{\left(\frac{1}{2}\right)}^2 \\ s=\frac{3}{2} \\ s=1.5 \end{gathered}$$

Hence, option C is correct.