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a car starts from rest at a stop sign. It accelerates for 4.0 m/s2 for 6.0 s, coasts for 2.0s, and then slows down at a rate of 3.0m/s2 for the next stop sign. How far apart are the stop signs?

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Distance travelled during first acceleration , S1 = ut1 + 1/2 at12 = 0*6 + 1/2*4*62 = 72 m

Speed after this phase of acceleration, v = u + at1 = 0 + 4*6 = 24 m/s

Distance travelled during coasting, S2 = v*t2 = 24*2 = 48 m

Distance travelled during deceleration, S3 = v2-u2/2a = (0-242)/(-2*3) = 96 m

Thus total distance = S1 + S2 + S3 = 72+48+96 = 216 m

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