Question

Two friends part company at 11pm. The first walks north at 1.2 m/sec, while the second walks east at 1.8 m/sec. How fast is the distance between them increasing 45 seconds later? Round your answer to the nearest tenth of a m/sec.

Answer 1

Let,

a= distance covered by 1^st person in north in time t.

b=distance covered by 2^nd person in east in time t.

Let x be the direct distance between both at time t.

And we can calculate this x using pythagoras theorem.

$\boxed {x=\sqrt{ a^2+b^2}}$

Here given,

a=1.2*45=54

b=1.8*45=81

Therefore,x=11.61

Now,

$\boxed {x^2={ a^2+b^2}}$

Now for to calculate h ow fast x is increasing, we will differentiate the function w.r.t x.

$2x{dx\over dt}=2a+2b\\$

Now at given a and b

$2x{dx\over dt}=2.88t+6.48t \\ 2x{dx\over dt}=9.36t$

Now at t= 45 and x=11.61

${dx\over dt}={9.36\over 2*(11.61)}$

${dx\over dt}=18.139$

After rounding off to nearest tenth,

$\boxed{{dx\over dt}=20m /s}$