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**Relative Motion**

**Ques 35:**** Two particles 1 and 2 are thrown in the directions shown in figure simultaneously with velocities 5 m/s and 20 m/s. Initially particle 1 is at height 20 m from the ground. Taking upwards as the positive direction, find:****(a) acceleration of 1 with respect to 2 ****(b) initial velocity of 2 with respect to 1 ****(c) velocity of 1 with respect to 2 after time t = 1/2 s ****(d) time when the particles will collide.****Ans: **(a) zero

(b) 25 ms^{-1}

(c) - 25 ms^{-1}

(d) 0.8 s**Sol:** (a) Acceleration of 1 w.r.t. 2

= (- g) - (- g)

= 0 m/s^{2}

(b) Initial velocity of 2 w.r.t. 1

= (+20) - (-5)

= 25 m/s

(c) Initial velocity of 1 w.r.t. 2

= (- 5) - (+ 20)

= - 25 m/s

âˆ´ Velocity of 1 w.r.t. 2 at time

(d) Initial relative displacement of 2 w.r.t. 1

as at time t (= time of collision of the particles) the relative displacement of 2 w.r.t. 1 will be zero

â‡’ t = 0.85 s**Ques 36: A person walks up a stalled 15 m long escalator in 90 s. When standing on the same escalator, now moving, the person is carried up in 60 s. How much time would it take that person to walk up the moving escalator? Does the answer depend on the length of the escalator?****Ans: **36 s, No**Sol:** Let length of escalator L (= 15 m)

walking speed of man = L/90

Speed of escalator = L/60

Time taken by man walking on a moving escalator

Required time has been found without using the actual length of the escalator.**Ques 37: A ball is thrown vertically upward from the 12 m level with an initial velocity of 18 m/s. At the same instant an open platform elevator passes the 5 m level, moving upward with a constant velocity of 2 m/s. ****Determine: ****(a) when and where the ball will meet the elevator, ****(b) the relative velocity of the ball with respect to the elevator when the ball hits the elevator.****Ans: **(a) 3.65 s, at 12.30 m level (b) 19.8 ms^{-1} (downwards)**Sol:**= Initial displacement of elevator w.r.t. ball = + 10 m.

Relative velocity of elevator w.r.t. ball

= (2) - (18) = - 16 m/s

Accelerator of elevator w.r.t. ball

= (0) - (- 10)

= 10 m/s^{2}

or 5t^{2} - 16t - 10 = 0

t = 3.65 s

Position of elevator when it meets ball

= 5 + (2 x 3.65)

= 12.30 m level**Ques 38: An automobile and a truck start from rest at the same instant, with the automobile initially at some distance behind the truck. The truck has a constant acceleration of 2.2 m/s ^{2} and the automobile has an acceleration of 3.5 m/s^{2}. The automobile overtakes the truck when it (truck) has moved 60 m. **

(b) 35.5 m

(c) automobile 25.9 ms

â€¦ (i)

â‡’ t = 7.39 s

â€¦(ii)

Dividing Eq. (ii) by Eq. (i),

or

x = 35.5 m

(c) At the time of overtaking

Speed of automobile = (3.5) (7.39)

= 25.85 m/s

Speed of truck = (2.2) (7.39)

= 16.25 m/s

âˆ´ acceleration of thrown body w.r.t. lift

= (- g) - (+ a)

= - (g + a)

If time of flight is t, using

v

(- v) = (+ u) + {- (a + g)}t

â‡’ (a + g) t = 2 u

or

(b) 50 m

Speed along OB is 2âˆš2 m/s

âˆ´ Time taken to reach B

= 10 s**Ques 41: Given = magnitude o f velocity o f boatman with respect to river, in the direction shown. Boatman wants to reach from point A to point B, At what angle Î¸ should he row his boat.****Ans: ****Sol: In Î” OPQ,****Ques 42: An aeroplane has to go from a point P to another point Q, 1000 km away due north. Wind is blowing due east at a speed of 200 km/h. The air speed of plane is 500 km/h. ****(a) Find the direction in which the pilot should head the plane to reach the point Q. ****(b) Find the time taken by the plane to go from P to Q.**** Ans: **(a) at an angle Î¸ = sin

** Sol: **Let pilot heads point R to reach point Q

â‡’ Î¸ = sin^{ -1} (0.4)

(PR)^{2} = (RQ)^{2} + (PQ)^{2}

or (500 t)^{2} - (200 t)^{2} = (1000)^{2}

or

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