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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 v1 and at time t =t is v2 . 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?
```
10 years ago

```							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)

```
10 years ago
```							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

```
10 years ago
```							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....

```
10 years ago
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