ILLUSTRATION : 25
On a two-lane road, car A is travelling with a speed of 36 km hl. Two cars B and C approach car A in opposite directions
with a speed of 54 km h-' each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to
overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
SOLUTION
Answers
Answered by
1
Answer:
Explanation:
Car A is in the middle. Car B and C are on either side of car A. Let car B travel in the same direction as A.
Va = speed of car A = 36 kmph,
Vb = - 54 kmph = Vc,
Relative velocity of c wrt A : 54+36 = 90 kmph,
distance between them = 1 km,
time to cross = 1/90 hrs,
Initial Relative velocity of car B wrt A : 54 - 36 = 18 kmph,
max. Time to cross = 1/90 hrs,
Distance : 1 km,
s = u t + 1/2 a t²,
1 = 18 / 90 + 1/2 * a /90²,
a = 2*90 * 72 *1000/(3600*3600) m/s^2 ,
a = 1 m/s^2 or 12,960 km/hour^2.
HOPE IT HELPS YOU.
Answered by
0
Answer:
Speed of A=36km/hr=36×
18
5
m/s=10m/s
Speed of B= Speed of C=54×
18
5
m/s=15m/s
Relative speed of A w.r.t C=10+15=25m/s
Time taken by C to overtake A=
25
1000
=40sec.
Distance travelled by A all this time =10×40=400m
So, B have to cover distance of (1000m+400m)
i.e. 1400m to take over a A before C does. in 40sec.
Now putting in formula;-
s=ut+
2
1
at
2
1400=15×40+
2
1
×a×(40)
2
or, 800=a×800
a=1m/s
2
Ans
Explanation:
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