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You just finished Kinematics

In this chapter, you have completed: 19 concepts 17 doubts More than 30 Questions and Examples In this chapter, firstly you read Concepts of “Kinematics” and then moved to “applications of kinematics”. Apart from this, there are certain tips for you which will be helpful for you to develop a lot of interest in this chapter. There are some examples as well as practise question in each...

A projectile is thrown at an angle \beta with vertical .It reaches a maximum height H . The time taken to reach the highest point of its path is 

@Harshit As we have learned Time of Flight - Time for which projectile remains in the air above the horizontal plane. - wherein SO  Time taken to reach highest point =       Time taken to reach highest point = 

Equation of trajectory of a projectile is given by y = -x^2 +10 x , where x and y are in metres and x is along horizontal and y is vertically upward and particles is projected from origin .Then which of the following option is incorrect 



Initial velocity of particle is \sqrt{505}m/s


horizontal range is 10 m 


Maximum range is 10 m 


Angle of projection with horizontal is \tan ^{-1} (-1 )

@Harshit Equation of path of a projectile - it is equation of parabola     Acceleratio due to gravity     initial velocity Angle of projection   - wherein Path followed by a projectile is parabolic is nature.     Compare this with    Maximum height H =       

A person A is moving along east and B is moving along north. Then relative velocity of A with respect to B is (V_{A}=10\, m/s\, ,\; V_{B}=10\sqrt{3}\, m/s)

    =    -      = 

A ball is thrown vertically upward with a velocity of 20 ms-1 from a top of a multistory  building.The hight of the  point from where  the ball is thrown is 25 m from the ground. The hight to which the ball rises from the ground and time taken before the ball hits the ground are (g:=10 ms-2)

  Maxmium height from the ground :20 + 25 = 45m Time to reach ground S=-25m, u=20 ms-1, a =-10 m/s2

The speed of a swimmer in still water is 20 m/s . The speed of river water is 10 m/s and is flowing due east . If he is standing on the south bank and wishes to cross he river along the sortest path , the angle at which he should make his strokes w.r.t  north is given by : 


  • Option 1)

    30 west 

  • Option 2)

    0 \degree

  • Option 3)

    60 \degree west

  • Option 4)

    45 \degree west

Option 1)30 west Option 2)Option 3)Option 4)

The radius of circle , the period of revolution , initial poisition and sense of revolution are indicated in fig 

y-projection of the radius vector of rotating particle P is

  • Option 1)

    y ( t ) = -3 \cos 2 \pi t , \: \: where \: \: y \: \: in \, \, m

  • Option 2)

    y ( t ) = 4 \sin \left (\frac{ \pi t}{2} \right ) , \: \: where \: \: y \: \: in \, \, m

  • Option 3)

    y ( t ) = 3 \cos\left ( \frac{3 \pi t }{2} \right ) , \: \: where \: \: y \: \: in \, \, m

  • Option 4)

    y ( t ) = 3 \cos\left ( \frac{\pi t }{2} \right ) , \: \: where \: \: y \: \: in \, \, m

At t =  0 , position is at 3 cm  and  Option 1)Option 2)Option 3)Option 4)

The x and y coordinates of the particle at any time are x = 5t - 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is

dx/dt =Vx and dy/dt = Vy a = dV/dt in x dirextion a = -4 m/s^2 and in y direction a =0
Screenshot_2019-03-12-10-32-15-75.png How that equation inversed??
If S1=S2=S3=S Then 

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v=\beta x^{-2n} where b and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

@Vishal gond  as  use  and   so 
A particle moves on a straight line in such a way that it covers first half distance with speed 3m/sec and next half distance in 2 equal time intervals with speed 4.5m/s and 7.5m/s respectively. Find the average speed of the particle.

  let total distance of path is S for first half , time taken if t speed = distance/time 3 = S/2t t = S/6      .............1 for second half , time taken is T ... for first  T/2 it moves with 4.5 speed = distance/time 4.5 = 2x/T  x = 4.5T/2   ...........2 for second T/2 it moves with 7.5 7.5 = 2y/T y = 7.5T/2       ..........3 we have , x+y = S/2so (7.5+4.5)(T/2) = S/2 T = S/12     ...
Two balls are thrown simultaneously,A vertically upward with speed 20m/s from the ground and B vertically downward from a height of 40m with same speed and along the same line motion. At which point ball will collide ??
X be the distance from the ground where the collision takes place then Distance from top of tower up to the collision point =40-X let collision took place after time t then X=20t-1/2×g×t^2……(1)(upward motion) 40-X=20t+1/2g×t^2….(2)(downward motion) Adding (1) and (2) we get 40=40t Which gives t=1 second Putting value of t in (1) we get  ( taking :g= 9.8 m /s^2) X=15.1 metre So collision takes...
if a stone is released from a 30 m high tower and at same time a stone with 25 m/s thrown vertically upward from the base of that tower then find the position at which both stones will meet and time taken by them to meet
@deeksha Relative distance=30 m Relative speed= 25 m/s Relative acceleration=0 SO use  so  Now for particle released from a 30 m high tower a=-g m/s2 and u=0 t=1.2 so use  both stones will meet  will meet 7.2 m below from the top of tower         
IMG_20190128_121547.jpg 1 if v=x square find relation between x-t if x=1 t=0 where x position t time v velocity

if v=x2   find relation between x-t if x=1 t=0 where x position t time v velocity

@deeksha and  so we get  put for x=1 , t=0 we get C=1 so 
A particle starts with initial velocity u = 10 m/s and acceleration given by a = t - 4 . Then its velocity after 4 second will be
actually it is 2 m/s
Screenshot_20180903-181845.jpg A particle moves so that its position vector is given by F = cos wt i + sin wt j Where w' is a constant. Which of the following is true?
  For two perpendicular vectors and  So, velocity is  to r and  is directed towards the origin. cos wt i —w sin wti — cos wti dt

And also say how to handle "how long will it take " type of questions.

Please answer these 3 questions.

4. A car along a straight highway with speed 126 km h-l is brought to a halt within a distance of 200m, What is the retardation of the car (assumed uniform) and how long does it take for the car to stop? 5. A car is moving with speed u Driver of the car sees red traffic light, His reaction time is t, then find out the distance travelled by the car after the instant when the driver decided to apply brakes 6.Assume uniform retardation 'a' after applying breaks. If a body starts from rest and travels 120cm in the 6th second then what is the acceleration

s we can find using


and t from v=u+at

where t=reaction time + time taken to stop


 A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity of sound, the time T after which the splash is heard is given by


Let's say time taken by stone to reach top of the well  is equal to t1 Now t2 is time taken by sound to reach the observer total time T = t1+t2  

 Sir ji help in this question and in video lecture u will directly  calculate the accleretion value but sir there are many other students like me who will have only subjects like P.C.B. So they are not undestand so kindly tell us how we are able to solve this types of questions in detail and and clear my above doubt also.

Assuming initial velocity is zero, so u = 0. hence   option (c) is correct


A particle p is projected from a point on the surface of a smooth inclined plane (see fig), simultaneously another particle Q isreleased on the smooth inclined plane from the same position. p and Q collide after t=4sec. The speed of  projection of  P is

@Arvind gupta  For Q U=0,   So  Now for P if the angle of projection of P with the incline surface is α  => ucosα = speed along the plane, a = gsinθ° = acceleration along the plane  Similarly usinα = speed perpendicular  to the plane, a = gcosθ° = acceleration perpendicular to the plane  time of flight (For P) = 2usinα/gcosθ° = 4 s  usinα = 2gcosθ° distance travelled down incline = s = ut +...