You've probably seen the problem where two people, AA and BB, walk with different constant speeds v1v_1 and v2v_2, starting at different positions on the number line p1p_1 and p2p_2, and then meeting at an unknown time. We solved the problem by solving the linear equation v1t+p1=v2t+p2v_1 t + p_1 = v_2 t + p_2, where tt is the unknown time.

Notice that this is equivalent to finding a zero for the function of their distance with each other f(t)=(v1−v2)t+(p1−p2)f(t) = (v_1 - v_2) t + (p_1 - p_2). Suppose that v1<v2v_1 < v_2 and p1>p2p_1 > p_2. Then f(0)=p1−p2>0f(0) = p_1 - p_2 > 0 and f(2(p1−p2)v2−v1)=(v1−v2)2(p1−p2)v2−v1+(p1−p2)=−(p1−p2)<0.f(\frac{2(p_1 - p_2)}{v_2 - v_1}) = (v_1 - v_2) \frac{2 (p_1-p_2)}{v_2 - v_1} + (p_1 - p_2) = -(p_1 - p_2) < 0. So based on IVT, there must be a time tt in (0,2(p1−p2)v2−v1)\left(0, \frac{2 (p_1-p_2)}{v_2-v_1}\right) such that they meet.

Now suppose AA and BB start walking from rest with accelerations a1a_1 and a2a_2 (with a1<a2a_1 < a_2), holding all other conditions the same. Then the equation we'll solve will be 12a1t2+p1=12a2t2+p2\frac{1}{2} a_1 t^2 + p_1 = \frac{1}{2} a_2 t^2 + p_2. Using IVT, during which one of the following time periods will they definitely meet?

Choose one answer

Type a concept, like “slope” or “photosynthesis”.

↑ ↓ to choose · Enter to openSearch explanations and problems