E-1.A car is moving towards south with a speed of 20 m/s. A motorc fli

1.	E-1.A car is moving towards south with a speed of 20 m/s. A motorc flist is moving towards east with aspeed of 15 mis. At a certain instant (t = 0) the motorcyclist is due south of the car and is at a distanceof 50m from the car. The shortest distance between the motorcyclist and the car is and the time afterWhich 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» Answer: 30m and 1.6sec 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

E-1.A car is moving towards south with a speed of 20 m/s. A motorc flist is moving towards east with aspeed of 15 mis. At a certain instant (t = 0) the motorcyclist is due south of the car and is at a distanceof 50m from the car. The shortest distance between the motorcyclist and the car is and the time afterWhich 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»

Answer:

30m and 1.6sec

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