A blue car of length 4.55 m is moving north on a roadway that intersects another

Question

A blue car of length 4.55 m is moving north on a roadway that intersects another perpendicular roadway (see figure below). The width d of the intersection from near edge to far edge is 25.0 m. The blue car has a constant acceleration of magnitude 1.60 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 3.00 s

(b) For what time interval is any part of the blue car within the boundaries of the intersection?

(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 5.30 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?

(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?

Explanation / Answer

(b) Initial velocity of the blue car when it enter into the intersection = u
time taken to cover the distance d = 25 m is t = 3 s
we know that
d = ut + (1/2)at2 where a is acceleration
here a = -1.6 m/s2 , now putting all the values
25 = u*3 - 0.5*1.6*32
u = 10.733 m/s
Now blue car has to cover the distance = d + 4.55
to get entirely out of the intersection
hence total distance S= 25 +4.55 = 29.55 m
29.55 = uT + (1/2)aT2
Where T is the total time of interval in which blue car will be in the intersection
29.55 = 10.733T - 0.5*1.6*T2
T = 3.87 s
(c) Red car start from rest , therfore (u) = 0
time to reach the intersection = time taken by the blue car to leave the intersection = 3.87 s
acceleration (a) = 5.3 m/s2
Hence
S = ut + (1/2)at2
S = 0*3.87 + (1/2)*5.3*3.872 = 39.69 m
(d)Speed after time t = 3.87 s
V = u +at = 0 +5.3*3.87 = 20.511 m/s2

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