At time t=0 Automobile A has velocity of 24mi/hr and Automobile B has velocity 3

Question

At time t=0 Automobile A has velocity of 24mi/hr and Automobile B has velocity 36mi/hr and both are travelling on adjacent highway lanes. Knowing that automobile A has constant acceleration of 1.8ft/s^2 and that B has constant deceleration of 1.2ft/s^2, Determine when and where A will overtake B and the speed of each automobile at that time. My attempt at a solution: I know I need to use the equation VB/A=VB-VA and that both units need to be converted to be similar. I found relative velocities and relative accelerations, but my numbers aren't coming out right. I also know that 12mi/hr=17.6 ft/s and the correct answer for t=~15 seconds

Explanation / Answer

vA0 = 24 mi/hr, vB0 = 36 mi/hr, aA = 1.8 ft/s2, aB = -1.2 ft/s2, find t, sand vA, vB

sA = va0t + aAt2/2 = vB0t + aBt2/2

so t = 2(vB0 - vA0)/(aA - aB) = 2*(12 mi/hr)/(3.0 ft/s2) = 2*17.6/3 = 11.7 s

s = vA0t + aAt2/2 = (24 mi/hr)*11.7 s + (1.8 ft/s2)*(11.7 s)2/2 = (2*17.6 ft/s)*11.7 s + 1.8*11.72/2 ft = 535 ft

vA = vA0 + aAt = 24 mi/hr + 1.8*11.7 ft/s = 2*17.6 ft/s + 1.8*11.7 ft/s = 56.3 ft/s

vB = vB0 + aBt = 36 mi/hr - 1.2*11.7 ft/s = 3*17.6 ft/s - 1.2*11.7 ft/s = 38.8 ft/s

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