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 \[a,b\] and \[c\] are real numbers greater than \[1\]. hen what is the average speed of the car over the time interval \[\left[ {{t_1},{t_2}} \right]\] is attained at the point?
A. \[\dfrac{{{t_1} + {t_2}}}{2}\]
B. \[2a\left( {{t_1} + {t_2}} \right) + b\]
C. \[\dfrac{{{t_2} - {t_1}}}{2}\]
D. \[a\left( {{t_2} - {t_1}} \right) + b\]

Answer 1

Hint: Find the position of the car at time \[{t_1}\] and \[{t_2}\]. Then find the difference of these two positions to get the distance travelled by the car over the time interval \[\left[ {{t_1},{t_2}} \right]\]. After that divide the distance by the required time \[\left( {{t_2} - {t_1}} \right)\]. You’ll get the average speed of the car over that time interval. Now, differentiate the given function to get the speed of the car at any instant. Let the average speed is attained at the point \[t'\]. Then equate the average speed and the speed at time \[t'\] and find the value of \[t'\], which is required.

Formula used
Average speed of a moving body over the time interval \[\left[ {{t_1},{t_2}} \right]\] is \[\dfrac{{f\left( {{t_2}} \right) - f\left( {{t_1}} \right)}}{{{t_2} - {t_1}}}\], where \[f\left( t \right)\] is the position of the body at time \[t\].
Speed of a moving body at time \[t\] is \[f'\left( t \right)\], where \[f\left( t \right)\] is the position of the body at time \[t\].
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] and derivative of any constant is zero.

Complete step by step solution
Given that \[f\left( t \right) = a{t^2} + bt + c\], \[t > 0\]
The average speed of the car over the time interval \[\left[ {{t_1},{t_2}} \right]\] is obtained by dividing the distance covered by the total time.
\[f\left( t \right)\] is the position of the moving car at time \[t\].
\[\therefore \]The car covers the distance \[f\left( {{t_1}} \right)\] at time \[{t_1}\] and \[f\left( {{t_2}} \right)\] at time \[{t_2}\].
So, the distance covered over the time interval \[\left[ {{t_1},{t_2}} \right]\] is \[\left| {f\left( {{t_2}} \right) - f\left( {{t_1}} \right)} \right|\]
And the time required is \[\left( {{t_2} - {t_1}} \right)\], since \[{t_2} > {t_1}\]
\[\therefore \] The average speed of the car over the time interval \[\left[ {{t_1},{t_2}} \right]\] is \[\dfrac{{f\left( {{t_2}} \right) - f\left( {{t_1}} \right)}}{{{t_2} - {t_1}}}\]
Let us find \[f\left( {{t_1}} \right)\] and \[f\left( {{t_2}} \right)\]
, so
\[f\left( {{t_1}} \right) = a{t_1}^2 + b{t_1} + c\] and \[f\left( {{t_2}} \right) = a{t_2}^2 + b{t_2} + c\]
\[\therefore f\left( {{t_2}} \right) - f\left( {{t_1}} \right)\]
\[ = \left( {a{t_2}^2 + b{t_2} + c} \right) - \left( {a{t_1}^2 + b{t_1} + c} \right)\]
\[ = a{t_2}^2 + b{t_2} + c - a{t_1}^2 - b{t_1} - c\]
\[ = a{t_2}^2 - a{t_1}^2 + b{t_2} - b{t_1} + c - c\]
\[ = a\left( {{t_2}^2 - {t_1}^2} \right) + b\left( {{t_2} - {t_1}} \right)\]
\[ = a\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right) + b\left( {{t_2} - {t_1}} \right)\]
\[ = \left( {{t_2} - {t_1}} \right)\left\{ {a\left( {{t_2} + {t_1}} \right) + b} \right\}\]
\[\therefore \dfrac{{f\left( {{t_2}} \right) - f\left( {{t_1}} \right)}}{{{t_2} - {t_1}}} = \dfrac{{\left( {{t_2} - {t_1}} \right)\left\{ {a\left( {{t_2} + {t_1}} \right) + b} \right\}}}{{\left( {{t_2} - {t_1}} \right)}} = a\left( {{t_1} + {t_2}} \right) + b\]
Given that \[f\left( t \right) = a{t^2} + bt + c\]
Differentiating \[f\left( t \right)\] with respect to \[t\], we get
\[f'\left( t \right) = \dfrac{d}{{dt}}\left( {a{t^2} + bt + c} \right)\]
\[ = a\dfrac{d}{{dt}}\left( {{t^2}} \right) + b\dfrac{d}{{dt}}\left( t \right) + \dfrac{d}{{dt}}\left( c \right)\]
Use the formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] and derivative of any constant is zero.
\[ \Rightarrow f'\left( t \right) = a \times 2t + b \times 1 + 0 = 2at + b\]
This is the speed of the car at the time \[t\].
Now, if the average speed is attained at time \[t'\], then
\[a\left( {{t_1} + {t_2}} \right) + b = 2at' + b\]
\[ \Rightarrow 2t' = {t_1} + {t_2}\]
\[ \Rightarrow t' = \dfrac{{{t_1} + {t_2}}}{2}\]
So, the average speed of the car is attained at the point \[\dfrac{{{t_1} + {t_2}}}{2}\].

Hence option A is correct.

Note: We get the speed of a body at any instant after differentiating the given function but to get average speed in a time interval you have to divide the distance traveled at that time interval by the required time. Here average speed is required. So, at first, the distance has been calculated, and then it has been divided by the total time.