1. The first friend is initially at a distance of 75.0m from the second friend. 2. The first friend is walking towards the second friend at a constant speed of 0.72 m/s. 3. The second friend is accelerating from rest at an average acceleration of 1.10 m/s^2. 4. We need to find the time it takes for the two friends to shake hands. 5. To solve this problem, we can use the equations of motion. 6. The equation for the distance traveled by an object with constant acceleration is given by: d = ut + (1/2)at^2, where d is the distance, u is the initial velocity, a is the acceleration, and t is the time. 7. We can set up two equations, one for each friend, and solve them simultaneously to find the time at which they meet. solve for t in 75 = 0.72t + (1/2)(1.10)t^2=Results: t = -2/55 (18 + sqrt(103449)) STEP 2: 1. I have the expression for t = -2/55 (18 + sqrt(103449)). 2. The time it takes for the two friends to shake hands is given by the positive value of t, since time cannot be negative. Final Answer: The two friends will shake hands approximately 18 + sqrt(103449) seconds after they start moving.