Question

Solution

Correct option is As shown in Fig. in time tA will move a distance v1t while Bv2t; so after time t the distance and B from  will be (l1 – v1t) and (l2 – v2t) respectively. So the distance between them L at time t will be given by  Differentiating Eqn. (1) with respect to time, Now for L to be minimum (dL/dt) = 0;  Substituting this value of t from Eqn. (2) in (1) and simplifying, we get Eqns. (2) and (3) are the desired results.

SIMILAR QUESTIONS

Q1

A particle moves along a straight line such that its displacement at any timet is given by s = (t3 + 6t2 + 3t + 4)m. What is the velocity of the particle when its acceleration is zero?

Q2

A particle starts moving from the position of rest under a constant acc. If it travels a distance x in t sec, what distance will it travel in next t sec?

Q3

A particle moving with velocity equal to 0.4 m/s is subjected to an acceleration of  0.15 m/s2 for 2 sec in a direction at right angle to its direction of motion. What is the magnitude of resultant velocity?

Q4

A body starting from rest slides on an inclined plane of length s as shown in Fig. Calculate the time of descent and speed at the bottom. Find also

(a)  Distance covered in half the time of descent and (b) time taken to cover half the distance Q5

A juggler throws balls into air. He throws one whenever the previous one is at its highest point. How high do the balls rise if the throws n balls each sec? Acceleration due to gravity is g.

Q6

pebble is thrown vertically upwards from a bridge with an initial velocity of 4.9m/s. It strikes the water after 2s. If acc. Due to gravity is 9.8m/s2 (a) what is the height of the bridge?(b) with what velocity does the pebble strike the water?

Q7

body is released from a height and falls freely towards the earth. Exactly 1 sec later another body is released. What is the distance between the two bodies 2 sec after the release of the second body, if g = 9.8m/s2?

Q8

If a body travels half its total path in the last second of its fall from rest, find; (a) the time and (b) the height of fall. Explain the physically unacceptable solution of the quadratic time equation. (g = 9.8m/s2)

Q9

stone is dropped into a well and the sound of impact of stone on the water is heard after 2.056 sec of the release of stone from the top. If acc. Due to gravity is 980cm/sec2 and velocity of sound in air is 350m/s, calculate the depth of the well.

Q10

person aims a gun at a target located at a horizontal distance of 100 m. If the gun imparts a horizontal speed of 500 ms-1 to the bullet, at what height above the target must he aim his gun in order to hit it? Take g = 10 ms-2.