Menu div.question,div.steps {font-size: 16px; padding: 10px;line-height: 20px; margin: 20px;} div.question{border: 1px solid #ccc;} div#solution .button{font-size: 20px; margin: 20px; width: 200px; padding: 10px; text-align: center;} div.hint,div.answer,div.steps{display: none;} div.pre_next{font-size: 30px; width:90%; margin:auto;} div.pre_next a.nav_pre{float:left;} div.pre_next a.nav_next{float:right;} div.title div.notes{ font-family:sans-serif, Arial; font-size: 18px; } 3. At t = 0, one toy car is set rolling on a straight track with initial position 15.0 cm, initial velocity -3.50 cm//s, and constant acceleration 2.40 cm//s^2. At the same moment, another toy car is set rolling on an adjacent track with initial position 10.0 cm, initial velocity +5.50 cm//s, and constant acceleration zero. (a) At what time, if any, do the two cars have equal speeds? (b) What are their speeds at that time? (c) At what time(s), if any, do the cars pass each other? (d) What are their locations at that time? (e) Explain the difference between question (a) and question (c) as clearly as possible. Hint x_f = x_i + v_it + 1/2 at^2 quadratic formula: ax^2 + bx + c = 0 than x = (-b ± sqrt(b^2 - 4ac))/(2a) Answer (a) 3.75\ s (b) 5.50\ cm//s (c) 0.604\ s,6.90\ s (d)13.3\ cm,47.9\ cm Show Steps Given: 1st car x_(i1) = 15cm v_(i1) = -3.5 cm//s a_1 =2.4 cm//s^2 2nd car x_(i2) = 10cm v_(i2) = 5.5 cm//s a_2 =0 (a) v_1 = v_(i1) + a_1t => t = (v_1 - v_(i1))/a_1 v_(i2) = 5.5 cm//s and keep same speed over time. v_1 = v_2 = 5.5 cm//s We get: t = (5.5 cm//s - (-3.5 cm//s))/(2.4 cm//s^2) = color(red)(3.75s) (b) When t=3.75s v_1 = -3.5 cm//s + 2.4 cm//s^2 times 3.75 s = color(red)(5.5 cm//s^2) (c) 1st car x_1 = x_(i1) + v_(i1)t + 1/2a_1t^2 = 15 - 3.5t -1.2t^2 2nd car x_2 = x_(i2) + v_(i12)t = 10 + 5.5t The passing point is where x_1 = x_2. We get: 15 - 3.5t -1.2t^2=10 + 5.5t =>1.2t^2 -9t +5 =0  Solving for t by using the quadratic formula: t =(9 ± sqrt(9^2 - 4times1.2times5))/(2times1.2) color(red)(t=6.9s) or color(red)(t=0.604s) (d) At 6.9s: x_1 = x_2 = 10.0 cm + 5.5m//s times 6.9s =color(red)(47.9 cm) At 6.9s: x_1 = x_2 = 10.0 cm + 5.5m//s times 0.604s =color(red)(13.3cm) (e) At first, car_1 start form 15.0 cm moving backward with a forward acceleration. car_2 start form 10.0cm moving forward with a constant speed. And 2 cars meet at x = 13.3cm. Then, car_1 stop and change direction with same acceleration. Finally, car_1 catch up the car_2 at x = 47.9m with a higher speed.