JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 10)
The position of a moving car at time t is
given by f(t) = at2 + bt + c, t > 0, where a, b and c are real numbers greater than 1. Then the average speed of the car over the time interval [t1 , t2 ] is attained at the point :
given by f(t) = at2 + bt + c, t > 0, where a, b and c are real numbers greater than 1. Then the average speed of the car over the time interval [t1 , t2 ] is attained at the point :
$${{\left( {{t_1} + {t_2}} \right)} \over 2}$$
$${{\left( {{t_2} - {t_1}} \right)} \over 2}$$
2a(t1
+ t2) + b
a(t2
ā t1) + b
Explanation
Vav = $${{f\left( {{t_2}} \right) - f\left( {{t_1}} \right)} \over {{t_2} - {t_1}}}$$ = f'(t)
$$ \Rightarrow $$ $${{a\left( {t_2^2 - t_1^2} \right) - b\left( {{t_2} - {t_1}} \right)} \over {{t_2} - {t_1}}}$$ = 2$$a$$t + b
$$ \Rightarrow $$ a(t2 + t1) + b = 2at + b
$$ \Rightarrow $$ t = $${{{t_1} + {t_2}} \over 2}$$
$$ \Rightarrow $$ $${{a\left( {t_2^2 - t_1^2} \right) - b\left( {{t_2} - {t_1}} \right)} \over {{t_2} - {t_1}}}$$ = 2$$a$$t + b
$$ \Rightarrow $$ a(t2 + t1) + b = 2at + b
$$ \Rightarrow $$ t = $${{{t_1} + {t_2}} \over 2}$$
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