Question

Car A is parked along the northbound lane of a highway, and car B is traveling in the southbound lane at a constant speed of 60 mi/h. At t = 0, A starts and accelerates at a constant rate a_A, while at t = 5 s, B begins to slow down with a constant deceleration of magnitude a_A/6. Knowing that when the cars pass each other x = 294 ft and v_A = v_B, determine (a) the acceleration a_A, (b) when the vehicles pass each other, (c) the distance d between the vehicles at t = 0.

o-60 mi/h Fig. P11.45

Answer 1

Solution: UB)o = 60 1111 /h A/0 At position 1: t-0: VA ад constant VB--60 mi/hr- -88 ft/s constant XB At position 2: t = 5: At position 3: time from l to2: 5 S time from 2 to 3: t-5 XA=XB = 294 ft velocities are equal: va3--ViB Principles: Uniform acceleration and uniform velocitv 0 0 2 Equations of Motion for Car A between position 1 and 3: 2 294 -(a)t АЗ Equation of Motion of Car B between positions 1 and 2: B2 B2 Equation of Motion of Car B between positions 2 and 3: 2

294 X2(-88) (t -5) (a6)(t - 5) Solve the set of 6 equations identified on the previous page: Note that there are 6 unknowns: VB3, VA3, XB2, aa, t, and d A3-V B3 294- ()t2 X82 -d - 440 Start by eliminating XB2 and vB3 A3d 294()t2 294 (d - 440) (-88) (t 5)(a/6) (t - 5) next get rid of va3 and notice that d is only present in the final equation, so solve it for d a^t 5a 75.42 5 294 (a)t2 d 294+440 (88) (t -5)-(a/6)(t - 5) 294 -(a^)( 588a 0.5102a2 +107.67aA+ 5685.2 combine the first equations to solve for aA 75.42 107.67 5689.4 +0.714)2 -a.5102+ 0 0.5102a480.33a 5685.2 (solve with quadratic equation) Since 929.46 is very large, the Quadratic Formula correct acceleration value is 0.5102 -480.33 5685 a- aA-11.98 ft/s2 which give a total travel time of Radical: REb*b-4ac 75.4 11.98 + 0.714-7.008s 219114.9609 Solution: the distance d is d-294 + 440 + (88)(t-5) _ _ (aA /6)(t-5)2 929.4660627 11.98826895 two real roots

d - 294 + 440 + (88) (7-5) (12/6)(7 - 5)2 2 - 906 ft