Question

The position of a moving car at time t is given by $f\left(t\right)=a{t}^2+bt+c,t>0$, where $a,bandc$ are real numbers greater than $1$. Then the average speed of the car over the time interval $[t_1,t_2]$ is attained at the point

• A. $\frac{(t_2+t_1)}{2}$
• B. $2a(t_1+t_2)+b$
• C. $\frac{(t_2-t_1)}{2}$
• D. $a(t_2-t_1)+b$

Answer 1

The correct option is A

$\frac{(t_2+t_1)}{2}$

Step 1: Given Data:

Equation$f\left(t\right)=a{t}^2+bt+c,t>0$

Real numbers$=a,bandc$

Step 2: Formula used:

Velocity is defined as the rate of change of acceleration with respect to time

$v=\frac{df}{dt}$

Step 3: Calculating the average speed of the car:

Using the formula, the average speed can be calculated as-

$\begin{array}{rcl}f’\left(t\right)& =& V_{av}\\ & =& \frac{f\left(t_2\right)-f\left(t_1\right)}{(t_2-t_1)}\\ & =& \frac{[a({t^2}_2-{t^2}_1)+b(t_2-t_1)]}{(t_2-t_1)}\\ & =& a(t_1+t_2)+b\end{array}$

Also,

$$\begin{gathered} v=\frac{d\left(a{t}^2+bt+c\right)}{dt} \\ v=2at+b \end{gathered}$$

Now, equate

$\begin{array}{rcl}2at+b& =& a\left(t_1+t_2\right)+b\\ t& =& \frac{\left(t_1+t_2\right)}{2}\end{array}$

Hence option A is the correct answer.