Question

The two roads AB and BC intersect at the point B, where ABC = 60 ° and AB = 28 m. One person starts cycling from point A to B at a speed of 4 meters per second and at that time another person starts cycling from point B at a speed of 4 meters per second on BC road. What is the rate at which the distance between them changes after 3 seconds of travel?​

Answer 1

Step-by-step explanation:

BP=28–4t ; BQ=8t ; Let PQ=x

By cosine rule x2=BP2+PQ2–2BP∗BQcos60∘=(28–4t)2+64t2–2∗(28–4t)∗8t∗0.5….(1)

Differentiating w.r.t. t

2xdxdt=8(28–4t)+128t−8(28–8t)

At t=3 , x=(28–4t)2+64t2–2∗(28–4t)∗8t−−−−−−−−−−−−−−−−−−−−−−−−−√=162+64∗9–8∗16∗8∗3−−−−−−−−−−−−−−−−−−−−√=21.16601m

Plugging t=3 in (2) 2∗21.16601dxdt=8∗16+384–32=480

dxdt=11.33893m/s

The distance is increasing at the rate of 11.33893

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Answer 2

Step-by-step explanation:

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