Question

a car on the cross
road is moving towards east at a speed of 40 km per hour another car which is 1 km south of the crossing is approaching the crossing with same speed their closest distance are approach will be

Answer 1

Answer:

1/√2 km

Explanation:

Given A car on the cross  

road is moving towards east at a speed of 40 km per hour another car which is 1 km south of the crossing is approaching the crossing with same speed their closest distance are approach will be

We know that tan θ = v1 / v2

     So θ = tan^-1 40/40

             = tan^-1 1

           θ  = 45 degree

Now for closest distance we have

Distance d = x sin θ

                 = 1 sin 45

                 = 1 x 1 / √2

            d = 1/√2 km

Answer 2

Answer:

0.707 km

Explanation:

A car on the cross road is moving towards east at a speed of 40 km/hr. Another car which is 1 km south of the crossing is approaching the crossing

with the same speed. Their closest distance of approach will be

Distance of Car A  from crossing =  40 T    km

Distance of Car B  from crossing = 1 - 40T  km

Distance between Car A & Car B =  √ ((40T)² + (1 - 40T)²)

= √ 1600T² + 1 + 1600T² - 80T

= √ (3200T² - 80T + 1)

d(D)/dt  =  (1/2) ( 6400T - 80)/(√ (3200T² - 80T + 1)

d(D)/dt  = 0

=> T = 80/6400

=> T = 1/80 hr

=> 40T = 1/2

Min Distance between Car A & B =  √ ((1/2)² + (1 - 1/2)²)

= √ (1/4 +1/4)

= √1/2

= 0.707 km