Question

E-1.
A car is moving towards south with a speed of 20 m/s. A motorc flist is moving towards east with a
speed of 15 mis. At a certain instant (t = 0) the motorcyclist is due south of the car and is at a distance
of 50m from the car. The shortest distance between the motorcyclist and the car is and the time after
Which they are nearest to each other after t=0
(1) 10 m, 180 (2) 20 m, 10
(3) 20 m, 164. (4) 40 m, 1 s.​

Answer 1

Answer:

30m

Explanation :

If "t" is the time at that point where the car is moving south due east.

Taking distnce moved towards south by car and motorciclist as 20t and 15t

Then by using pythagoras theorem,

distance, x=d^2=(50-20t)^2 - 15t^2

       

Thus, differentiating with respect to time t

         dx/dt = -40(50-20t) - 450t

Now, to check maxima and minima -

since in the question minimum distance is asked hence the dedifferentiation

must be >0

d^2x/dt^2 = 1250t

As t cannot be 0 hence >0 thus it is minima thus least distance exists.

Now to calculate time t,

dx/dt=0

On solving we get,

1250t = 2000

t=1.6sec

Now, on putting the value of t in the equation we get x=900.

So, d^2=900

shortest dist d= 30 m.