A blue car of length 4.37 m is moving north on a roadway that intersects another

Question

A blue car of length 4.37 m is moving north on a roadway that intersects another perpendicular roadway (see figure below). The width of the intersection from near edge to far edge is 29.5 m. The blue car has a constant acceleration of magnitude 1.90 m/s2 directed south. The time interval required for the nose of the blue car to move from the near (south) edge of the intersection to the north edge of the intersection is 3.20 s

(c) A red car is at rest on the perpendicular intersecting roadway. As the nose of the blue car enters the intersection, the red car starts from rest and accelerates east at 5.50 m/s2. What is the minimum distance from the near (west) edge of the intersection at which the nose of the red car can begin its motion if it is to enter the intersection after the blue car has entirely left the intersection?
m

(d) If the red car begins its motion at the position given by the answer to part (c), with what speed does it enter the intersection?

Part c is correct, only need help with part d.

Explanation / Answer

similar solution (a) Lenght of the blue car, L=4.62 m width of the road,d=30 m acceleration of the blue car,aB=-1.60 m/s2 accelration of the red car, aR=5.30 m/s2 time intervael for nose of the blue car to cross inter section, t=3.30 s Let, VoB be the velocity of the blue car on reaching the intersetion from equation of motion, d=VoBt + 1/2aBt2 30 m=VoB(3.3 s)+ 1/2(-1.60m/s2)(3.3 s)2 VoB=6.45 m/s The distance form the blue car from the south edge is form equation of motion, V2-VoB2 =2as 02-(6.45 m/s)2=2(-1.6m/s2)(s) s=13 m ________________________________________________________________ ________________________________________________________________ (b) total distance traveld, s' =d+L =30 m + 4.62 m =34.62 m Time intervel for any part of blue car within the boundiries T s'=VoB T + 1/2 aB T2 (34.62 m)=(6.45 m/s) T + 1/2(-1.60 m/s2) T2 (0.8 m/s2) T2 -(6.45 m/s) T + (34.62 m) = 0 T=3.68 s ________________________________________________________________ ________________________________________________________________ (c) intial velociyof the red car, VoR = 0 m/s distance of the red car formthe wst edge, dR from th ewquation of motion, DR = 0 T + 1/2 (5.30 m/s2)(T)2 =1/2 (5.30 m/s2)(3.68 s)2 =35.88 m _____________________________________________________________ _____________________________________________________________ (d) speed of the red car on entering the intersection (VR) from the equation of motion, VR2-VoR2 =2aB DR 02-VoR2 =2(-1.60 m/s2)(35.88 m) VoR =10.71 m/s

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