1).Determine the uniform acceleration of A and B: A is ____ft/s^2. B is ____ft/s

Question

1).Determine the uniform acceleration of A and B: A is ____ft/s^2. B is ____ft/s^2.
2).Determine when the vehicles pass each other: ____s.
3).Determine the speed of B when A and B pass each other. B is ____mi/h. Thank you for your help! Two automobiles A and Bare approaching each other in adjacent highway lanes. At t 0, A and Bare 3200 ft apart, their speeds are vA E 66 milh and VB E 39 mi/h, and they are at points Pand Q, respectively. A passes point Q 40 s after B was there and B passes point P42 s after A was there. 3200 ft References eBook & Resources Section Break Problem 11.043 Uniformly accelerated motion of two particles-DEPENDENT MULTIPART PROBLEM ASSIGN ALL PARTS Required information 10.00 points Problem 11.043-a Uniformly accelerated motion of two particles Determine the uniform accelerations of A and B. (Round the final answer to three decimal places.) (You must provide an answer before moving to the next part.) The uniform acceleration of A is fts2 The uniform acceleration of Bis ft/s

Explanation / Answer

here,

vA = 66 mi/h = 29.5 m/s

vb = 39 mi/h = 17.43 m/s

d = 3200 ft = 975.36 m

1)

let the accelration of car A be aA

time taken , t1 = 40 s

using seccond equation of motion

d = vA * t1 + 0.5 * aA * t1^2

975.36 = 29.5 * 40 + 0.5 * aA * 40^2

aA = - 0.26 m/s^2 = - 0.85 ft/s^2

let the accelration of car B be aB

time taken , t2 = 42 s

using seccond equation of motion

d = vB * t2 + 0.5 * aB * t2^2

975.36 = 17.43 * 42 + 0.5 * aB * 42^2

aB = 0.304 m/s^2 = 1 ft/s^2

2)

let they passes each other at time t and distance x from A

for A

x = vA * t+ 0.5 * aA * t^2

x = 29.5 * t - 0.5 * 0.26 * t^2 ....(1)

and for B

(d - x) = vB * t + 0.5 * aB * t^2

(975.36 - x) = 17.43 * t + 0.5 * 0.304 * t^2 ....(2)

from (1) and (2)

t = 20.58 s

they will meet after 20.6 s

3)

the speed of B when they meet , V = vb + aB * t

V = 17.43 + 0.304 * 20.58 = 23.69 m/s

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