A car starts from rest at a stop sign. It accelerates at 4.6 m/s^2 for 6.2 s , coasts for 2.1s , and then slows down at a rate of 3.3 m/s^2 for the next stop sign. How far apart the are the stop signs?

Answer:D = 271.54 mExplanation:given,1. car accelerates at 4.6 m/s² for 6.2 s2. constant speed for 2.1 s3. slows down at 3.3 m/s² distance travel for case 1using equation of motion  [tex]d_1 = u t +\dfrac{1}{2}at^2[/tex]  [tex]d_1 =\dfrac{1}{2}\times 4.6\times 6.2^2[/tex]       d₁ = 88.41 mcase 2 constant speed for 2.1 s now, we have to find velocityv = u  + atv = 0 + 4.6 x 6.2v = 28.52 m/sdistance travel in case 2d₂ = v x td₂ = 28.52 x 2.1 = 59.89 mfor case 3 distance travel by the car v² = u² + 2 a sfinal velocity if the car is zero0² = 28.52² + 2 x (-3.3) x d₃6.6 d₃ = 813.39  d₃ = 123.24 mtotal distance travel by the car D = d₁ + d₂ + d₃D = 88.41 + 59.89 + 123.24D = 271.54 m