# Q ; A particle is projected vertically upward with an initial velocity 40 m/s. Find the displacement and distance covered by particle in 6 seconds . Take g = 10 m/s Answer given is 60 m , 100 m My Answer coming is 60 m , 90 m Q :A particle moves in straight line with uniform acc. its vel. at time t=0 is v 1 and at time t =t is v 2 . the avg vel. of particle in this time interval is ? Q:A particle moves in a straight line with const. acc..If it covers 10 m in first second and extra 10 m in next second.Find its initial velocity?

**
**

**Q ; A particle is projected vertically upward with an initial velocity 40 m/s. Find the displacement and distance covered by particle in 6 seconds . Take g = 10 m/s**

Answer given is 60 m , 100 m

My Answer coming is 60 m , 90 m

Q :A particle moves in straight line with uniform acc. its vel. at time t=0 is v_{1} and at time t =t is v_{2} . the avg vel. of particle in this time interval is ?

Q:A particle moves in a straight line with const. acc..If it covers 10 m in first second and extra 10 m in next second.Find its initial velocity?

## 3 Answers

Dear Shobhit,

This is a very straight forward question

at t = 0, u=40

since gravity is in the opposite direction as of velocity, it will retard the motion and a time will come when its upward velocity will become 0.

to calculate that time, use 1st equation of motion

v = u + a*t, a = -10 ,

0 = 40 - 10 * t

=> t = 4 s

at t = 4 s particle will start coming down

Distance covered till 4 s ,

v^2 = u^2 + 2* a * s

s= 40*40/2*10

s= 80 m

from t=4 to t=6 , it will come down

s1=0.5*a* t^2

s1=0.5*10*2*2

s1=20m

Displacement = 80-20 = 60m

Distance = 80 + 20 = 100m(as it covers 20m coming down)

For Q2

avg. vel.= (v1+v2)/2

since acc is uniform its avg. vel can be calculated very easily

for Q3

Let its initial velocity = u

and acc.= a

velocity after 1 sec = u + a

now

10= u + 1/2*a*1^2(after 1 s)

for next sec,

20 = u+1/2*a*2^2

Taking difference of these equations, we get

a = 20/3

and putting this value of a in 1st eqn , we get u=20/3 m/s

for q1. jst cheqout aftr hw mch tym dus d vel bcum 0:

given u=40

v=0; g=10

using v=u-gt

u get t=4 sec.......i.e.aftr 4 sec d vel bcums zero n d tym givn in ques is 6 sec'wich means dat d particl went up and aftr 4 sec it cam to rest den again gaind acc n came down...........so displacement can b calc. using s=ut-0.5g(t*t)

d dist covrd will be (dist cov b4 cumin 2 rest ) + (dist cov in 6-4=2 sec)........

n u'll gt d rite ans...

q2. Avg vel=net displacement divided by total time taken

acc=(v2-v1)/t

u=v1

v=v2

calc 's'

total tym is t.....

hence calc avg vel.

Q3.u cn solve d ques using 2 eq.

for 1st eq.

s=10; t=1; u and a unknown;

s=ut+0.5a(t*t)

for 2nd eq.

s=10; t=1; a remains same and u=vel attaind aftr 1 sec i.e.(u+at) and t=1

solve d 2 eq. simultaneously n u'll gt u and v both....