Question

Questions 1 &2

This is a more challenging problem with constant acceleration. Simply plugging numbers into an equation gets us nowhere, so something extra is required. On a movie set, a red car and a green car are to move toward each other on opposite sides of a highway lane divider line, as indicated in the figure. Green car Example 2.4.4 Figure 1 When the drivers pass each other, the driver of the red car is to toss a package of contraband to the other driver. To catch the toss on video, one of the cameras must be set up near the point where the toss will be made. When "action" is yelled out on the set, the initial separation between the drivers will be 200 m, the red car will accelerate from rest at a constant 6.12 m/s2, starting at x = 0, and the green car will already be in motion with a constant speed of 60 km/h at x, = 200 m. e far from the

Answer 1

0212 Xp= 3.06+p2 - 0 aps 200-16.667 tp -@ Equate both the values 3.06 tp 2 - 200 -16.667 tp 3.06 tp? +16.667 tp - 200 = 0 tp = -16.667 I 116.667)+ 4x3.06 x 200 2x3.06 to - 16 -7 + 277.74 + 2448 6:12 take tres .-16.687 +52:21 -5.815 6.12 so xp = 3.06 (5.8122 lap = 103.21m itp=5.815