Question

answer the following true or false: two cars drive from one spotlight to the next, leaving at the same time and arriving at the same time. then there is a time when they are going the same speed. true false

Explanation:

Step1: Define functions

Let \(s_1(t)\) and \(s_2(t)\) be the position functions of the two cars. Let \(t = a\) be the starting - time and \(t = b\) be the arrival - time. So \(s_1(a)=s_2(a)\) and \(s_1(b)=s_2(b)\).

Step2: Apply the Mean Value Theorem

By the Mean Value Theorem, the velocity of the first car \(v_1=\frac{s_1(b)-s_1(a)}{b - a}\) and the velocity of the second car \(v_2=\frac{s_2(b)-s_2(a)}{b - a}\). Since \(s_1(a)=s_2(a)\) and \(s_1(b)=s_2(b)\), we have \(v_1 = v_2\). Also, if we consider the difference function \(h(t)=s_1(t)-s_2(t)\), \(h(a)=h(b) = 0\). By Rolle's theorem (a special case of the Mean Value Theorem), \(h^\prime(c)=0\) for some \(c\in(a,b)\), where \(h^\prime(t)=v_1(t)-v_2(t)\) (since \(v(t)=s^\prime(t)\)).

Answer:

True