Two cars A and B are at rest at reference line of straight two lane road. A starts at t = 0 with constant acceleration of 1m/s^2 and attains a maximum velocity of 8 m/ s and moves with constant velocity. Car B starts at t = 4s and wants to overtake A in minimum time. Maximum speed and acceleration of B are 12 m/s and 3m/s^2 .
Answers
Answer:
Distance traveled by car A=S
A
Distance traveled by car B=S
B
According to formula S=ut+
2
1
at
2
S
A
=0t+
2
1
×2t
2
S
B
=10t+
2
1
×0t
2
Car A will overtake car B when
S
A
=S
B
+100
Putting the value S
A
and S
B
+2=10t+100
+2−10t−100=0
After solving for t, we get
t=16.18sec possible
t=−6.10sec not possible
So option B is correct.
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Option B) is correct
Explanation:
This question states that there are two cars named "A" and "B". Both these cars are at rest on a straight line. At time t = 0, car "A" starts with an acceleration of 1 m/s². After starting it attains a velocity of 8 m/s, and then it goes with a constant velocity. Now after 4 seconds, car "B" starts moving from rest at t = 4, with acceleration 3 /s², and the speed of car "B" is 12 m/s. Thus, if car "B" wants to overtake car "A" in minimum time it should take a lead of 12 m. Putting on the formula s = ut + 1/2at², we attain two values, one is 16.18m and another is -6.81. Since we exclude negative value, so the answer is 16.18 m, hence option B) is correct.