Question

1) At 2.00 pm, a car’s speedometer reads 30 mi/h. At 2.10 pm it reads 50mi/h. Show that at some time between 2.00 pm and 2.10 pm the acceleration is exactly 120 mi/h²

2) Two runners start a race at the same time and finish in a tie. Prove that at some time during the race they have the same speed. [Hint: Consider f(t)=g(t)-h(t) ,where g and h are the position functions of the two runners.]

Answer 1

Answer on Question #77374 – Math – Algebra

1) $v = v_{0} + a * \frac{1}{6}$ ; $a * \frac{1}{6} = 50 - 30 = 20$ ; $a = 120$ .

Answer: $a = 120$.

2) It is given a function $f(t) = g(t) - h(t), t \in [0, n]$ ,

$$f (t _ {0}) = f (t _ {n}) = 0$$

If runners move with the same speed all the time, then it is true that they have the same speed during the race. If the speed of runners differ, because when one of them increased distance to another one, his speed $(v_{1})$ was greater than the speed of the second runner $(v_{2})$ , namely $v_{1} > v_{2}$ . Whereas they finished together, then with a decline of $v_{1}$ , its value was approaching to $v_{2}$ , until at a certain point of time $v_{1} = v_{2}$ .

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