Inactive Tutor answered 06/09/19
In these types of questions, always ask yourself, "What's the same for both cars?" In this case, the distance they travel is the same. So, we'll apply the equation Δx = vot + 1/2at2. Both cars start from rest, so it reduces to Δx = 1/2at2. They have the same distance, but different accelerations and different times. So, ΔxA = 1/2aAtA2 = ΔxB = 1/2aBtB2. Set 1/2aAtA2 = 1/2aBtB2. Dividing both sides by 1/2 and substituting accelerations, 3.5tA2=4.9tB2. Car B is moving for 1 second less than Car A, so tB=tA-1. Let t = tA, 3.5t2 = 4.9(t-1)2. I can use algebra to solve for t: divide both sides by 4.9, so 0.714t2=(t-1)2, take the square root of both sides, so 0.845t=t-1, 1=0.155t, t=6.45s or t=6.458s if you don't round until the end. Alternatively, I can graph both equations for Δx and find the intersection, arriving at the same answer, t=6.458s.