The driver of a 2.0 × 103 kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car in front of him, which had come to rest because of blocking ahead as shown in above Fig. After the brakes are applied, a constant friction force of 7.5 × 103 N acts on the car. Ignore air resistance. (a) Determine the least distance should the brakes be applied to avoid a collision with the other vehicle?
Answers
Answer:
Answer:
Initial speed of the car A u=52 kmh
−1
=52×
18
5
=14.44 ms
−1
The car stops in 5 seconds i.e v=0 at t=5
(A) : Initial speed of the car A u=3 kmh
−1
=3×
18
5
=0.83 ms
−1
The car stops in 10 seconds i.e v=0 at t=10
With the help of these initial and final points for both the cases, we can plot the graph of speed vs time.
Area under the speed-time graph gives the distance covered.
∴ Distance covered by car A x
A
=
2
1
×14.44×5=36.1 m
∴ Distance covered by car A x
B
=
2
1
×0.83×10=4.15 m
Thus car A travels more distance than B.
Explanation:
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