Why Kaysons ?

Video lectures

Access over 500+ hours of video lectures 24*7, covering complete syllabus for JEE preparation.

Online Support

Practice over 30000+ questions starting from basic level to JEE advance level.

Live Doubt Clearing Session

Ask your doubts live everyday Join our live doubt clearing session conducted by our experts.

National Mock Tests

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

Organized Learning

Proper planning to complete syllabus is the key to get a decent rank in JEE.

Test Series/Daily assignments

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

SPEAK TO COUNSELLOR ? CLICK HERE

New Basic Physics

  • 1
  • 2
  • 3
  • 4

Q.

The displacement  of a particle moving in one dimension under the action of a constant force is related to time  by the equation , where  is in metres and  is in seconds. Find the displacement of the particle when its velocity is zero                    

Q.

A particle moves along a straight line such that its displacement  varies with time  as 

Column-I

Column- II

  1. Acceleration at  s
  2. Average velocity during 3rd sec
  3. Velocity at   s
  4. Initial displacement

p.  

 r.  

 

s.  

t.  

Q.

The velocitytime relation of an electron starting from rest is given by where . The distance traversed in first 3 s is 

Q.

A particle is moving along the -axis whose acceleration is given by , where  is the location of the particle. At , the particle is at rest at  m. The distance travelled by the particle in 5 s is        

Q. An object moves along the x−axis. Its x−coordinates is given as a function of time as x=7t−3t 2 m where x is in meters and t is in seconds. Its average speed over the interval t=0 to t=4 s is?

X=7t−3t 2 ⇒v= dt dx​=7−6t Let us find the time when the velocity becomes zero Putting v=0⇒7−6t=0 ⇒t=7/6 s x (t−7/6 s)​=7× 6 7​−3× 6 2 7 2​= 12 49​m At t=0→x=0 At t=4 s→x=7×4−3(4) 2 =−20 m Distance travelled =s 12 49​+ 12 49​+20= 6 169​ms −1

Q. The relation between time t and displacement x is t = αx^2+βx, where α and β are constants. The retardation is I have successfully differentiated till velocity but from differentiating acceleration why we are using d(x^m + N)^n/dx identity but why not d(x+N) ^n/dx, since the x in velocity is linear.?

you will get velocity in terms of x , then to find accleration you can use chain rule formula

a = v*dv/dx

Q. what is meant by potential gradient?

potential gradient dV / dx is difference in potential between two points, or per unit length

Q. In acceleration how to use a= v*dv/ds and In which kind of example? And what is the difference between a = v*dv/ds and a = dv/dt??

accleration is the rate of change in velocity

a = dv / dt

now using chain rule

a = dv / ds * ds /dt [ where ds is the change in displacement and ds / dt = velocity ]

so a = dv / ds * v

so when velocity is given as function of s , then we can use this formula

Q. An object moves along x axis. It's X coordinates is given as a function of time as x = 7t - 3tsquare.Find the average speed from t = 0 to t = 4s?

vavg = ∆x/ ∆t = x2 − x1 / t2 − t1

where x1(t = 0) = 0;

x2(t = 4) = 7(4) − 3(4)2 = −20 m

So; vavg = −20 − 0 / 4 − 0 = −5.00 m/s