NEET Physics Motion in A Straight Line Questions Solved


 

A ball is thrown upward with a certain speed. It passes through the same point at 3 second and 7 second from the start. The maximum height achieved by the ball is

1.  500 m

2.  250 m

3.  125 m

4.  450 m

(3)                    

As upward and downward motion are symmetric [Time taken from AB is same as time taken from BC]

So time to reach highest point is 5s. Same time is required to reach from B to ground.

Time to reach highest point after 3sec = 7-32=2sec.

Time to reach highest point from ground = 3+2 = 5sec.

For downward motion,

u=0a=gt=5sh=12gt2=12×10×52=125 m

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a=dvdt=dvds×dsdt=v×dvds

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Difficulty Level:

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(1)

 

Motion of A wrt B

Relative velocity of A wrt B = VA - VB                                                =v- (vcos60° )                                                 = v/2

 Time =av2=2av

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Distance travelled in t second,

Here, initial velocity u=0h=gt22                                                         ......        (1)Distance travelled in nth second = u +a2(2n -1)Distance travelled in last second= 0 + g2(2t-1) =9h25     ..............(2)Equating h from equation 1 and 225g(2t-1)18=gt229t2 -50t +25 =0Solving for t,t =50 ±502-4×9×252×9 = 1018s or 5 sTime has to be greater than 1 second, so  t=5sPutting value of t in eqn. 1h=g522 = 122.5 m

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A point moves in a straight line under the retardation av2. If the initial velocity is u, the distance covered in 't' seconds is-

1.  aut

2.  1aln aut

3.  1aln 1+aut

4.  a ln aut

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(2)

 

Upward journey                                                                   Downward journey

                                      

Net acceleration = -(a+g)                                          Net acceleration = g - a
-ve indicates retardation

 

tv= 2hg+atD= 2hg-atUTD= g-ag+a= 812= 23

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